[proofplan] The forward direction is pointwise: each tangent vector is a derivation at its base point, so applying the pointwise product rule at every $p\in M$ gives the Leibniz rule for $D_X$. For the converse, we first define a derivation at each point by $X_p(f)=D(f)(p)$ and show it depends only on the germ of $f$ at $p$ using a bump function. Locality then lets us extend coordinate functions to global smooth functions and compute the local coordinate coefficients of $X$ as $D$ applied to those extensions. Since those coefficients are smooth functions, the field is smooth, and uniqueness follows from equality of the actions on all smooth functions. [/proofplan]

[step:Verify that a smooth vector field acts as a derivation]Let $X\in\mathfrak{X}(M)$ be a smooth vector field. Define \begin{align*} D_X:C^\infty(M)&\to C^\infty(M) \end{align*} by \begin{align*} D_X(f)(p)=X_p(f) \end{align*} for $f\in C^\infty(M)$ and $p\in M$. Since $X$ is a smooth vector field, the function $p\mapsto X_p(f)$ is smooth for every $f\in C^\infty(M)$, so $D_X(f)\in C^\infty(M)$. For $a,b\in\mathbb{R}$ and $f,g\in C^\infty(M)$, the $\mathbb{R}$-linearity of each tangent derivation $X_p$ gives \begin{align*} D_X(af+bg)(p)=X_p(af+bg)=aX_p(f)+bX_p(g)=aD_X(f)(p)+bD_X(g)(p). \end{align*} Thus $D_X$ is $\mathbb{R}$-linear. For $f,g\in C^\infty(M)$ and $p\in M$, the Leibniz rule for the tangent vector $X_p$ gives \begin{align*} D_X(fg)(p)=X_p(fg)=f(p)X_p(g)+g(p)X_p(f). \end{align*} Since $X_p(g)=D_X(g)(p)$ and $X_p(f)=D_X(f)(p)$, this becomes \begin{align*} D_X(fg)(p)=\bigl(fD_X(g)+gD_X(f)\bigr)(p). \end{align*} Because this holds for every $p\in M$, \begin{align*} D_X(fg)=fD_X(g)+gD_X(f). \end{align*}[/step]

[guided]Let us check the forward direction directly from the definition of tangent vectors as derivations. A smooth vector field assigns to each point $p\in M$ a tangent vector $X_p\in T_pM$, and a tangent vector at $p$ acts on smooth functions by an $\mathbb{R}$-linear derivation at $p$. Thus, for a fixed smooth function $f\in C^\infty(M)$, the formula \begin{align*} D_X(f)(p)=X_p(f) \end{align*} defines a real number at every $p\in M$. The smoothness condition in the definition of a smooth vector field says precisely that the resulting function $p\mapsto X_p(f)$ is smooth. Hence $D_X(f)\in C^\infty(M)$. Now fix $a,b\in\mathbb{R}$ and $f,g\in C^\infty(M)$. At each point $p\in M$, the tangent vector $X_p$ is $\mathbb{R}$-linear as a map on smooth functions, so \begin{align*} D_X(af+bg)(p)=X_p(af+bg)=aX_p(f)+bX_p(g). \end{align*} Using the definition of $D_X(f)$ and $D_X(g)$, this is \begin{align*} D_X(af+bg)(p)=aD_X(f)(p)+bD_X(g)(p). \end{align*} Since the equality holds at every $p$, the functions are equal, and $D_X$ is $\mathbb{R}$-linear. For the product rule, take $f,g\in C^\infty(M)$. At a fixed point $p\in M$, the tangent vector $X_p$ satisfies the pointwise Leibniz rule \begin{align*} X_p(fg)=f(p)X_p(g)+g(p)X_p(f). \end{align*} Substituting the definition of $D_X$ gives \begin{align*} D_X(fg)(p)=f(p)D_X(g)(p)+g(p)D_X(f)(p). \end{align*} The right-hand side is exactly the value at $p$ of the smooth function $fD_X(g)+gD_X(f)$. Therefore \begin{align*} D_X(fg)=fD_X(g)+gD_X(f). \end{align*}[/guided]

[step:Define the tangent vector determined by a derivation at each point] Let \begin{align*} D:C^\infty(M)\to C^\infty(M) \end{align*} be an $\mathbb{R}$-[linear map](/page/Linear%20Map) satisfying \begin{align*} D(fg)=fD(g)+gD(f) \end{align*} for all $f,g\in C^\infty(M)$. For each $p\in M$, define \begin{align*} X_p:C^\infty(M)&\to\mathbb{R} \end{align*} by \begin{align*} X_p(f)=D(f)(p). \end{align*} The map $X_p$ is $\mathbb{R}$-linear because $D$ is $\mathbb{R}$-linear. For $f,g\in C^\infty(M)$, \begin{align*} X_p(fg)=D(fg)(p)=f(p)D(g)(p)+g(p)D(f)(p). \end{align*} Hence \begin{align*} X_p(fg)=f(p)X_p(g)+g(p)X_p(f). \end{align*} Thus $X_p$ is a derivation at $p$, and therefore defines an element of $T_pM$ under the definition of tangent vectors as derivations at a point. [/step]

[step:Prove that the derivation is local at each point]Since $M$ is Hausdorff and second-countable, the smooth bump-function lemma applies: if $p\in M$ and $W\subset M$ is an open neighbourhood of $p$, then there exists $\chi\in C^\infty(M)$ such that $\chi(p)=1$ and $\operatorname{supp}\chi\subset W$; equivalently, after possibly shrinking around $p$, $\chi=1$ on a neighbourhood of $p$. Let $p\in M$, and let $h\in C^\infty(M)$ vanish on an open neighbourhood $W\subset M$ of $p$. Choose $\chi\in C^\infty(M)$ such that $\chi(p)=1$ and $\operatorname{supp}\chi\subset W$. Since $h=0$ on $W$ and $\chi=0$ on $M\setminus W$, we have $\chi h=0$ on all of $M$. Applying $D$ gives \begin{align*} 0=D(0)(p)=D(\chi h)(p). \end{align*} By the Leibniz rule, \begin{align*} D(\chi h)(p)=\chi(p)D(h)(p)+h(p)D(\chi)(p). \end{align*} Since $\chi(p)=1$ and $h(p)=0$, this yields \begin{align*} 0=D(h)(p). \end{align*} Thus, if a smooth function vanishes on a neighbourhood of $p$, then $D(h)(p)=0$. Consequently, if $f,g\in C^\infty(M)$ agree on a neighbourhood of $p$, then $f-g$ vanishes on a neighbourhood of $p$, and the preceding result gives \begin{align*} D(f)(p)-D(g)(p)=D(f-g)(p)=0. \end{align*} Therefore $D(f)(p)=D(g)(p)$ whenever $f$ and $g$ have the same germ at $p$.[/step]

[guided]The main issue in the converse is that $D$ is defined on global smooth functions, while tangent vectors are local objects. We need to show that $D(f)(p)$ only depends on the germ of $f$ near $p$. We use the standard smooth bump-function lemma on smooth Hausdorff second-countable manifolds: if $p\in M$ and $W\subset M$ is an open neighbourhood of $p$, then there is a smooth function $\chi\in C^\infty(M)$ with $\chi(p)=1$ and $\operatorname{supp}\chi\subset W$. This is the point where the usual smooth-manifold convention, together with the existence of smooth bump functions, enters the proof. Let $h\in C^\infty(M)$ vanish on an open neighbourhood $W$ of $p$. Choose such a bump function $\chi$ with $\operatorname{supp}\chi\subset W$ and $\chi(p)=1$. The product $\chi h$ is identically zero on $M$: on $W$ we have $h=0$, and outside $W$ we have $\chi=0$ because the support of $\chi$ is contained in $W$. Therefore \begin{align*} D(\chi h)(p)=D(0)(p)=0. \end{align*} The Leibniz rule for $D$ gives \begin{align*} D(\chi h)(p)=\chi(p)D(h)(p)+h(p)D(\chi)(p). \end{align*} Now $\chi(p)=1$, and $h(p)=0$ because $h$ vanishes near $p$. Hence \begin{align*} 0=D(h)(p). \end{align*} This proves locality in the vanishing case. If $f,g\in C^\infty(M)$ agree on a neighbourhood of $p$, then $h=f-g$ vanishes on a neighbourhood of $p$. By $\mathbb{R}$-linearity of $D$ and the vanishing case, \begin{align*} D(f)(p)-D(g)(p)=D(f-g)(p)=0. \end{align*} Thus \begin{align*} D(f)(p)=D(g)(p). \end{align*} So $D(f)(p)$ depends only on the germ of $f$ at $p$, which is exactly what will allow local coordinate functions to be used after extending them to global smooth functions.[/guided]

[step:Compute the local coordinate expression of the induced field] Let $(U,\varphi)$ be a coordinate chart on $M$ with coordinate functions \begin{align*} x_i:U&\to\mathbb{R} \end{align*} defined by \begin{align*} x_i(q)=\pi_i(\varphi(q)), \end{align*} where $\pi_i:\mathbb{R}^n\to\mathbb{R}$ is the $i$th coordinate projection and $n=\dim M$. For each $p\in U$, define the coordinate derivation $\partial_{x_i}\big|_p:C^\infty(M)\to\mathbb{R}$ by \begin{align*} \partial_{x_i}\big|_p(f)=\frac{\partial ((f|_U)\circ\varphi^{-1})}{\partial x_i}(\varphi(p)). \end{align*} This is the tangent vector at $p$ associated to the $i$th coordinate direction in the chart $(U,\varphi)$. For each $i\in\{1,\ldots,n\}$, define a function $a_i:U\to\mathbb{R}$ as follows. Given $p\in U$, choose an open neighbourhood $W\subset U$ of $p$ and a bump function $\chi\in C^\infty(M)$ such that $\chi=1$ on $W$ and $\operatorname{supp}\chi\subset U$. Define a smooth extension $\widetilde{x}_i:M\to\mathbb{R}$ by setting $\widetilde{x}_i=\chi x_i$ on $U$ and $\widetilde{x}_i=0$ on $M\setminus U$. Then $\widetilde{x}_i=x_i$ on $W$, and the extension is smooth because $\chi$ vanishes on a neighbourhood of $M\setminus U$. Set \begin{align*} a_i(p)=D(\widetilde{x}_i)(p). \end{align*} By the locality proved above, $a_i(p)$ is independent of the chosen neighbourhood, bump function, and extension. For every $f\in C^\infty(M)$, the tangent vector $X_p$ has the coordinate expression \begin{align*} X_p(f)=\sum_{i=1}^n a_i(p)\,\partial_{x_i}\big|_p(f). \end{align*} Indeed, the coordinate functions $(x_1,\ldots,x_n)$ form a coordinate system near $p$, and the previous step showed that $X_p$ is a derivation at $p$. Hence the coordinate-basis theorem for tangent derivations in a chart applies: every derivation $v\in T_pM$ has the form \begin{align*} v(f)=\sum_{i=1}^n v(x_i)\,\partial_{x_i}\big|_p(f) \end{align*} for all smooth functions $f$ defined near $p$, after replacing the local coordinate functions by any global smooth extensions. Applying this to $v=X_p$, the coefficient of $\partial_{x_i}\big|_p$ is its value on the $i$th coordinate function. Here that value is \begin{align*} X_p(\widetilde{x}_i)=D(\widetilde{x}_i)(p)=a_i(p), \end{align*} where locality justifies using the global extension $\widetilde{x}_i$ because $\widetilde{x}_i=x_i$ near $p$. [/step]

[step:Show that the local coefficients are smooth] It remains to prove that the coefficients $a_i$ vary smoothly. Let $p\in U$. Choose an open neighbourhood $W\subset U$ of $p$ and global smooth extensions \begin{align*} \widetilde{x}_i:M&\to\mathbb{R} \end{align*} such that $\widetilde{x}_i=x_i$ on $W$. For every $q\in W$, locality gives \begin{align*} a_i(q)=D(\widetilde{x}_i)(q). \end{align*} Since $D(\widetilde{x}_i)\in C^\infty(M)$ by the codomain of $D$, the restriction \begin{align*} a_i|_W=D(\widetilde{x}_i)|_W \end{align*} is smooth. Thus each coordinate coefficient $a_i$ is smooth locally on $U$, and hence smooth on $U$. Therefore, in the chart $(U,\varphi)$, \begin{align*} X|_U=\sum_{i=1}^n a_i\,\partial_{x_i}, \end{align*} with $a_i\in C^\infty(U)$. Since smoothness of a vector field is local in coordinate charts, $X$ is a smooth vector field on $M$. [/step]

[step:Identify the derivation and prove uniqueness] For every $f\in C^\infty(M)$ and every $p\in M$, the construction gives \begin{align*} D_X(f)(p)=X_p(f)=D(f)(p). \end{align*} Hence $D_X(f)=D(f)$ for every $f\in C^\infty(M)$, so $D_X=D$. Finally, suppose $Y\in\mathfrak{X}(M)$ is another smooth vector field with $D_Y=D$. Then for every $p\in M$ and every $f\in C^\infty(M)$, \begin{align*} Y_p(f)=D_Y(f)(p)=D(f)(p)=X_p(f). \end{align*} Thus $Y_p=X_p$ as derivations at $p$ for every $p\in M$. Therefore $Y=X$, proving uniqueness. [/step]