Let $F : M\to N$ be a map between manifolds. By definition the pushforward of a derivation $X$ in $p$ is $$dF_p(X)(f)=X(f\circ F)$$. Now if $M$ and $N$ are vector spaces, is it true that maybe under some identification, we have $dF_p(X) = \lim_{t\to 0}(F(p+tX)-F(p))/t$ (i.e. the directionnal derivative) ?
I don't really get how to show this. I wan't to show this because it because it seems more easy to handle in practical calculations.
The expression you've written is the identification, more or less.
For a finite dimensional real vector space $V$, there is a canonical isomorphism $l_v:V\to T_vV$ for each $v\in V$. Since it maps vectors onto derivations based at $v$, we can define it by defining how $l_v(u)$ acts on a smooth function $f\in C^\infty V$: $$ l_v(u)f=\frac{d}{dt}f(v+tu)|_{t=0}=\lim_{t\to 0}\frac{f(v+tu)-f(v)}{t} $$ This is precisely the identification one uses when taking directional derivatives in $\mathbb{R}^n$. In fact, more or less all of multivariable calculus can be rephrased in the language of manifolds by composing with these isomorphisms in order to describe everything in terms of the underlying vector space(s).
For a smooth map $F:U\to V$, the expression you've written down is precisely $l_{F(u)}^{-1}\circ d_uF\circ l_u:U\to V$.