Suppose we have a manifold $M$ and a Lagrangian function $L:TM\rightarrow \mathbb{R}$. I want to prove that a curve is a critical point of the action if and only if it satisfies the Euler-Lagrange equations.
So we have $S(c(s))=\int_{a}^{b}L(c(s,t),c'(s,t))dt$, and $\frac{d}{ds}S(c(s))|_{s=0}=\frac{d}{ds}\int_{a}^{b}L(c(s,t),c'(s,t))dt$ . Now all the proofs I have seen then go one to use the Leibniz rule and the derivative of the composition of functions and this is where I have my doubts on why we can use since $L$ is a function from the tangent bundle to $\mathbb{R}$. Now I guess there is something probably hapenning using charts that is not mentioned but I am not quite convinced and I can't seem to do with myself. Any help with this is aprecciated. Thanks in advance.
The standard derivation of the Euler-Lagrange equation on manifolds uses coordinates. Without coordinates (or some other splitting of $TTM$), one can't take a "horizontal derivative" analogous to $\partial L/\partial x$. The fact that the Lagrangian is written as a function of two arguments suggests that this is already being done in your case, at least implicitly.
Let $x^1,\cdots,x^n:U\to\mathbb{R}$ be a set of local coordinates on an open set $U\subseteq M$. These induce coordinates $(x^1,\cdots,x^n,v^1,\cdots,v^n)$ on $\pi^{-1}(U)\subseteq TM$ in the standard way. We can write the local representative of the Lagrangian in these coordinates as $L(x,v)=L(x^1,\cdots,x^n,v^1,\cdots,v^n)$.
Using local coordiantes, the derivation of the EL equation proceeds almost exactly like it does in $\mathbb{R}^n$, with the caveat that one must work locally. To show that a stationary path $c$ satisfies the EL-equations in local coordinates, it is sufficient to restrict attention to local variations such that $c$ only changes inside of (a compact subset of) the coordinate domain. For the converse, you can cover the path with coordinate charts and use a partition of unity.
Edit: Here's how to "localize" the first variation formula in a particular set of coordinates.
The action integral is perfectly well defined independently of coordinates $$ S(\gamma)=\int_a^bL(\dot{\gamma}(t))dt $$ Where $\gamma:[a.b]\to M$ is a curve and dots represent derivatives with respect to $t$. Given coordinates $x^i$ on $U$ which intersect with the image of $\gamma$, we may choose an interval $[a',b']\subseteq[a,b]$ with $\gamma([a',b'])\subset U$, and a variation $\gamma_s$ such that $\gamma_0=\gamma$ and $\gamma_s(t)=\gamma(t)$ for $t\notin(a',b')$. Before imposing coordinates, we may commute the integral and derivative $$ \frac{d}{ds}S(\gamma_s)|_{s=0}=\int_a^b\frac{d}{ds}L(\dot{\gamma}_s(t))|_{s=0}dt $$ However, since $\gamma_s$ is constant with respect to $s$ outside of $(a',b')$, we have $$ \frac{d}{ds}S(\gamma_s)|_{s=0}=\int_{a'}^{b'}\frac{d}{ds}L(\dot{\gamma}_s(t))|_{s=0}dt $$ and now, since $\gamma_s(t)\in U$ for $t\in[a',b']$, $s\in(-\epsilon,\epsilon)$, the integral can be passed to local coordinates without issue.