I am trying to understand the following: Let $\gamma: \mathbb{R} \supset I \to M$ be a smooth curve (M is a smooth manifold). Then the push forward $\gamma_{*}(\partial t)$ of $\partial t \in T_{t_0}\mathbb{R}$ is the tangent vector $\gamma'(t_0) \in T_{\gamma(t_0)}M$, where $\gamma'(t_0) \sim t \mapsto \gamma(t-t_0)$.
We defined the pushforward of a function $F:N \to M$ at the point $p$ as $F_{*}([\gamma]) = [F \circ \gamma]$, where $l \in [\gamma]$ if $\gamma(0)=l(0)=p$ and $\gamma_{\alpha}'(0) = l_{\alpha}'(0)$ for any (thus every) given chart $\phi_{\alpha}$ and $l_{\alpha}= \phi_{\alpha} \circ l$.
Now if I follow this definition, then $\gamma_{*}(\partial t) = \gamma_{*}([\partial t]) = [\gamma \circ \partial t]$. This should be easy, but I have no idea how to go on. Can anyone give me a hint?
From your notations, I am assuming that you are defining a tangent vector as an equivalence classes of curves. With this definition, $\partial_t \in T_{t_0}\mathbb R$ is the equivalence class of $t\mapsto t+ t_0$.
Given a smooth map $f:N\to M$ and a tangent vector $X = [\gamma] \in T_p N$, the pushforward is defined as the equivalence class of the composite $F\circ \gamma$ ie : $$f_*[\gamma] = [f\circ\gamma]$$
In our case, we have $\gamma:\mathbb R\to M$ and : $$\gamma_*\partial t = \gamma_*[t\mapsto t+t_0] =[t\mapsto \gamma(t+t_0)] = \gamma'(t_0)$$
NB : I believe there is a sign error in your post.