[proofplan]
We work in one coordinate chart containing the image of the solution on a subinterval. The invariance of the Lagrangian under the tangent-lifted point transformation is differentiated with respect to the symmetry parameter at $s=0$. This produces the infinitesimal invariance identity, and the Euler-Lagrange equations convert that identity into the total time derivative of the momentum pairing with the infinitesimal generator. Since this derivative vanishes, the Noether quantity is constant on the coordinate interval, and therefore locally constant along the solution.
[/proofplan]
[step:Pass to local coordinates along the solution]
Fix a coordinate chart $(U,x)$ and a subinterval $J\subset I$ such that $q(J)\subset U$. Write
\begin{align*}
a:J\to x(U),\quad t\mapsto x(q(t)).
\end{align*}
For each $i\in\{1,\dots,n\}$, let $a_i:J\to\mathbb{R}$ denote the $i$-th coordinate function of $a$, and write $\dot a_i(t)=\frac{d}{dt}a_i(t)$. Define the coordinate representative of the Lagrangian by
\begin{align*}
L_U:x(U)\times\mathbb{R}^n\to\mathbb{R},\quad (y,v)\mapsto L((d x^{-1})_y(v)).
\end{align*}
The coordinate representative of the infinitesimal generator is the smooth map
\begin{align*}
\zeta:x(U)\to\mathbb{R}^n,\quad y\mapsto d x_{x^{-1}(y)}(\xi(x^{-1}(y))).
\end{align*}
Write its components as $\zeta_i:x(U)\to\mathbb{R}$, so that $\zeta(y)=(\zeta_1(y),\dots,\zeta_n(y))$. The quantity to prove constant on $J$ is therefore
\begin{align*}
\mathcal{J}_U(t)=\sum_{i=1}^n \frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\zeta_i(a(t)).
\end{align*}
[/step]
[step:Differentiate the lifted symmetry in coordinates]
For $s$ sufficiently close to $0$, define the local coordinate representative of $\Phi_s$ by
\begin{align*}
F_s:x(U)\to\mathbb{R}^n,\quad y\mapsto x(\Phi_s(x^{-1}(y)))
\end{align*}
on the part of $x(U)$ where the expression is defined. Since $F_0(y)=y$, its infinitesimal generator is
\begin{align*}
\frac{\partial F_s}{\partial s}\Big|_{s=0}(y)=\zeta(y).
\end{align*}
The tangent lift of $\Phi_s$ sends the coordinate velocity vector $\dot a(t)\in\mathbb{R}^n$ to
\begin{align*}
\frac{d}{dt}F_s(a(t)).
\end{align*}
Thus the invariance hypothesis gives, for every $t\in J$ and every sufficiently small $s$,
\begin{align*}
L_U(F_s(a(t)),\frac{d}{dt}F_s(a(t)))=L_U(a(t),\dot a(t)).
\end{align*}
Differentiate this identity with respect to $s$ at $s=0$. By the chain rule in the finite-dimensional coordinates $(y,v)\in x(U)\times\mathbb{R}^n$,
\begin{align*}
0=\sum_{i=1}^n \frac{\partial L_U}{\partial y_i}(a(t),\dot a(t))\,\zeta_i(a(t))+\sum_{i=1}^n \frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\frac{d}{dt}\zeta_i(a(t)).
\end{align*}
[guided]
The point of passing to coordinates is that the tangent lift has a concrete expression. The point transformation is represented by the smooth map
\begin{align*}
F_s:x(U)\to\mathbb{R}^n,\quad y\mapsto x(\Phi_s(x^{-1}(y))).
\end{align*}
Because $\Phi_0=\operatorname{id}_Q$, we have $F_0(y)=y$. Differentiating $F_s$ with respect to $s$ at $s=0$ gives precisely the coordinate components of the infinitesimal generator:
\begin{align*}
\frac{\partial F_s}{\partial s}\Big|_{s=0}(y)=\zeta(y).
\end{align*}
Now evaluate the tangent-lift invariance on the velocity of the curve $q$. In coordinates, the curve is $a:J\to x(U)$, and its velocity is $\dot a(t)$. The tangent map of $F_s$ sends this velocity to the velocity of the transformed coordinate curve $t\mapsto F_s(a(t))$, namely
\begin{align*}
\frac{d}{dt}F_s(a(t)).
\end{align*}
Therefore the invariance of $L$ becomes the coordinate identity
\begin{align*}
L_U(F_s(a(t)),\frac{d}{dt}F_s(a(t)))=L_U(a(t),\dot a(t)).
\end{align*}
We now differentiate this identity with respect to the symmetry parameter $s$ at $s=0$. The first argument of $L_U$ contributes the term involving $\frac{\partial L_U}{\partial y_i}$, and the second argument contributes the term involving $\frac{\partial L_U}{\partial v_i}$. Since $\frac{\partial F_s}{\partial s}|_{s=0}(a(t))=\zeta(a(t))$ and differentiation in $s$ commutes with differentiation in $t$ for the smooth map $(s,t)\mapsto F_s(a(t))$, the velocity variation is
\begin{align*}
\frac{\partial}{\partial s}\Big|_{s=0}\frac{d}{dt}F_s(a(t))=\frac{d}{dt}\zeta(a(t)).
\end{align*}
Thus the differentiated invariance identity is
\begin{align*}
0=\sum_{i=1}^n \frac{\partial L_U}{\partial y_i}(a(t),\dot a(t))\,\zeta_i(a(t))+\sum_{i=1}^n \frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\frac{d}{dt}\zeta_i(a(t)).
\end{align*}
[/guided]
[/step]
[step:Use the Euler-Lagrange equations to identify a total derivative]
Since $q$ satisfies the Euler-Lagrange equations for $L$, its coordinate representative $a:J\to x(U)$ satisfies, for each $i\in\{1,\dots,n\}$,
\begin{align*}
\frac{d}{dt}\frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))=\frac{\partial L_U}{\partial y_i}(a(t),\dot a(t)).
\end{align*}
Substituting these identities into the infinitesimal invariance identity gives
\begin{align*}
0=\sum_{i=1}^n \frac{d}{dt}\frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\zeta_i(a(t))+\sum_{i=1}^n \frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\frac{d}{dt}\zeta_i(a(t)).
\end{align*}
By the product rule applied to each summand, the right-hand side is
\begin{align*}
\frac{d}{dt}\left(\sum_{i=1}^n \frac{\partial L_U}{\partial v_i}(a(t),\dot a(t))\,\zeta_i(a(t))\right).
\end{align*}
Hence
\begin{align*}
\frac{d}{dt}\mathcal{J}_U(t)=0
\end{align*}
for every $t\in J$.
[/step]
[step:Conclude local conservation of the Noether quantity]
The function $\mathcal{J}_U:J\to\mathbb{R}$ is differentiable because $L_U$, $\zeta$, and $a$ are smooth enough along the $C^2$ curve $q$. Since $\frac{d}{dt}\mathcal{J}_U(t)=0$ for every $t\in J$, $\mathcal{J}_U$ is constant on each connected component of $J$. This is exactly the asserted conservation law in the chosen coordinate chart. Because the chart and the subinterval were arbitrary, the Noether quantity is conserved locally along the solution wherever the coordinate expression is defined.
[/step]
