[proofplan] We compare the two coordinate expansions of the tangent vector by applying them to the $y_j$ coordinate function near $p$. In the $y$-coordinate basis, this application extracts the coefficient $w_j$. In the $x$-coordinate basis, the same application is computed by differentiating the $j$-th component of the transition map $\psi \circ \varphi^{-1}$. Equality of the two evaluations gives the coordinate transformation law. [/proofplan]

[step:Evaluate the tangent vector on a coordinate function in the $y$-chart]Fix $j \in \{1,\ldots,n\}$. Let \begin{align*} Y_j: V \to \mathbb{R} \end{align*} be the $j$-th coordinate function of the chart $\psi$, so $Y_j(q)=y_j(q)$ for $q \in V$. Let $\delta_{kj}$ denote the Kronecker delta, equal to $1$ when $k=j$ and equal to $0$ when $k \neq j$. By the definition of the coordinate basis vector $\partial_{y_k}\big|_p$ as the derivation differentiating the $k$-th coordinate direction in the chart $\psi$, \begin{align*} \partial_{y_k}\big|_p(Y_j)=\delta_{kj}. \end{align*} Applying the expansion \begin{align*} v=\sum_{k=1}^n w_k\,\partial_{y_k}\big|_p \end{align*} to the germ of $Y_j$ at $p$ gives \begin{align*} v(Y_j)=\sum_{k=1}^n w_k\,\partial_{y_k}\big|_p(Y_j). \end{align*} Substituting $\partial_{y_k}\big|_p(Y_j)=\delta_{kj}$ yields \begin{align*} v(Y_j)=w_j. \end{align*}[/step]

[guided]Fix $j \in \{1,\ldots,n\}$. The idea is to choose a [test function](/page/Test%20Function) that reads off exactly one $y$-coordinate component. Define \begin{align*} Y_j: V \to \mathbb{R} \end{align*} by declaring $Y_j(q)=y_j(q)$ for each $q \in V$. This is a smooth real-valued function on the chart domain $V$. Tangent vectors at $p$ act on germs of smooth functions defined near $p$, so it is legitimate to apply $v$ to $Y_j$ even though $Y_j$ is only defined on $V$, because $p \in V$. Let $\delta_{kj}$ be the Kronecker delta, meaning $\delta_{kj}=1$ if $k=j$ and $\delta_{kj}=0$ otherwise. The coordinate vector $\partial_{y_k}\big|_p$ is defined by differentiating in the $k$-th Euclidean coordinate after passing to the chart $\psi$. Since $Y_j \circ \psi^{-1}$ is the $j$-th coordinate projection on $\psi(V) \subset \mathbb{R}^n$, its $k$-th [partial derivative](/page/Partial%20Derivative) is $\delta_{kj}$. Therefore \begin{align*} \partial_{y_k}\big|_p(Y_j)=\delta_{kj}. \end{align*} Now apply the $y$-coordinate expansion of $v$ to this germ: \begin{align*} v(Y_j)=\left(\sum_{k=1}^n w_k\,\partial_{y_k}\big|_p\right)(Y_j). \end{align*} Linearity of the derivation gives \begin{align*} v(Y_j)=\sum_{k=1}^n w_k\,\partial_{y_k}\big|_p(Y_j). \end{align*} Using $\partial_{y_k}\big|_p(Y_j)=\delta_{kj}$, the sum reduces to \begin{align*} v(Y_j)=\sum_{k=1}^n w_k\,\delta_{kj}=w_j. \end{align*} Thus evaluating $v$ on the $j$-th $y$-coordinate function extracts precisely the $j$-th $y$-coordinate component of $v$.[/guided]

[step:Compute the same evaluation in the $x$-chart] Let \begin{align*} \Theta: \varphi(U \cap V) \to \psi(U \cap V) \end{align*} be the coordinate transition map \begin{align*} \Theta=\psi\circ\varphi^{-1}. \end{align*} Let \begin{align*} \Theta_j: \varphi(U \cap V) \to \mathbb{R} \end{align*} be its $j$-th component. Since \begin{align*} Y_j \circ \varphi^{-1}=\Theta_j \end{align*} on $\varphi(U \cap V)$, the definition of the coordinate basis vector $\partial_{x_i}\big|_p$ gives \begin{align*} \partial_{x_i}\big|_p(Y_j)=\frac{\partial \Theta_j}{\partial r_i}(\varphi(p)), \end{align*} where $(r_1,\ldots,r_n)$ denotes the standard coordinate system on $\mathbb{R}^n$. By the notation fixed in the statement, this is \begin{align*} \partial_{x_i}\big|_p(Y_j)=\frac{\partial y_j}{\partial x_i}(p). \end{align*} Applying the expansion \begin{align*} v=\sum_{i=1}^n v_i\,\partial_{x_i}\big|_p \end{align*} to $Y_j$ and using linearity gives \begin{align*} v(Y_j)=\sum_{i=1}^n v_i\,\partial_{x_i}\big|_p(Y_j). \end{align*} Therefore \begin{align*} v(Y_j)=\sum_{i=1}^n v_i\,\frac{\partial y_j}{\partial x_i}(p). \end{align*} [/step]

[step:Equate the two computations] The two preceding steps compute the same real number $v(Y_j)$. Hence \begin{align*} w_j=\sum_{i=1}^n v_i\,\frac{\partial y_j}{\partial x_i}(p). \end{align*} Since $j \in \{1,\ldots,n\}$ was arbitrary, the formula holds for every coordinate component. This proves the asserted change-of-coordinates rule for tangent vector components. [/step]