Let $a$ and $b$ in $M$, let $\gamma: [0, 1]\rightarrow M$ a path which is differentiable coordinates by coordinates.
Do we have $\dot{\gamma}(t)=(\frac{d}{dt}\gamma_1(t), ..., \frac{d}{dt}\gamma_n(t))\in T_{\gamma(t)}(M)$?
It seems trivial to me but I can't find this written anywhere explicitly, when I dig into it, it requires more knowledge about manifolds than I do have.
Yes. A priori a path is a smooth map $\gamma:I\rightarrow M$, where $I\subset \mathbb R$ is an open (or I guess closed) interval. If we define tangent space to be the set of derivations at a point $p\in M$, and similarly for a point $t_0\in I$, then the differential of a smooth map is given by: $$(D_{t_0}\gamma(v))(f)=v(f\circ \gamma)$$ where $f\in C^\infty(M)$, and $v\in T_{t_0}I$. It is easily checked this is a derivation at the point $\gamma(t_0)$ in $M$, so the differential is a smooth map between tangent spaces in this very abstract sense.
Since we now know that abstractly $D_{t_0}\gamma(v)\in T_{\gamma(t_0)}M$, we can calculate what this map is in coordinates. The coordinate representation of $D_{t_0}\gamma$ is just given by the Jacobian of $\phi\circ \gamma\circ \psi^{-1}$ for smooth charts $\psi$ and $\phi$ on $I$ and $M$ respectively. Since $I\subset \mathbb R$, we can just use the identity chart, hence if we have coordinates $\phi(p)=(x^1,\dots, x^n)$ on $M$, we can write $\gamma$ in coordinates as $\gamma(t)=(\gamma^1(t),\dots, \gamma^n(t))$, the Jacobian is then $\dot{\gamma}(t_0)=((\dot{\gamma}^1(t_0),\dots,\dot{\gamma}^n(t_0))$. Since the $1\in T_{t_0}I\cong \mathbb R$, we have that $$D_{t_0}\phi\circ\gamma(1)=((\dot{\gamma}^1(t_0),\dots,\dot{\gamma}^n(t_0))\in T_{\phi(\gamma(t_0)}\mathbb R^n$$, so the vector $\gamma^i(t_0)\frac{\partial}{\partial x^i}\Big|_{\gamma(t_0)}$ lies $T_{\gamma(t_0)}M$, which if you use the coordinates basis implicitly can be written as the row vector $((\dot{\gamma}^1(t_0),\dots,\dot{\gamma}^n(t_0))\in T_{\gamma(t_0)}M$