Computing the differential of a function on manifolds using velocity vector I am studying manifolds using Loring Tu's book and am slightly confused about the expression of the differential using the velocity vector. Below is a snippet of the book.

$T_pN$ is the tangent space of the manifold $N$ while $F_{*,p}$ is the differential. From what I understand, $F_{*,p}$ maps $X_p$ to an element in $T_{F(p)} M$, the tangent space of the manifold M at $F(p)$. However, the elements of the tangent spaces are derivations which maps $C^{\infty}(M)$ to $\mathbb{R}$. In other words, if $f \in C^{\infty}(M)$, then $F_{*,p}(X_p) f \in \mathbb{R}$.
Looking at this definition below in terms of the velocity, I am not sure where the "$f$" would go in here. Would it be the case that $F_{*,p}(X_p)f = \frac{d}{dt} \big|_0 (f \circ F \circ c)(t)$?
I am asking because the example presented after this proposition deals with the mapping $\ell_g: GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ given by $\ell_g(B) = gB$ for $g,B \in GL(n,\mathbb{R})$ (see snippet below). It then shows that the differential $(\ell_g)_*$ is also a left multiplication by $g$. Now, if we have $f \in C^{\infty}(GL(n,\mathbb{R}))$, the result derived by the example seems to suggest that $(\ell_g)_*(X) f = gX f$.
But, if I use my definition above, i.e.  ${\ell_g}_{*}(X)f = \frac{d}{dt} \big|_0 (f \circ \ell_g \circ c)(t)$, I don't seem to get $gXf$?
Clarifications would be appreciated!

 A: If $c$ is a curve on a manifold $M$ with $c(0)=p$ then there is a corresponding tangent vector $c'(0)\in T_p M$  which is defined by $c'(0)f:={(f\circ c\dot)}(0)$ for $f\in C_p^{\infty}M$. Here we use $()\dot{}$ for the calculus derivative whereas by $c'(0)$ we mean the point-derivation on $C_P^\infty M$.
If $M\subseteq \mathbb R^N$ is an open subset with the induced differentiable structure (here $M\subset\mathbb R^{n\times n}\approx \mathbb R^{n^2}$)
then for all $p\in M$ there is a well defined linear isomorphism $\Phi: T_pM \to \mathbb R^N$, $c'(0)\mapsto \dot c(0)$.
Now in our case if $X=c'(0)$ then
$$(l_g)_*(X)f=X(f\circ l_g)=c'(0)(f\circ l_g)=(f\circ l_g\circ c)\dot{}(0)=(l_g\circ c)'(0)f$$
so $(l_g)_*(X)=(l_g\circ c)'(0)$. Applying the identification $\Phi$ yields
$$\Phi((l_g)_*(X))=(l_g\circ c)\dot{}(0)=g\cdot \dot c(0)=g\cdot \Phi(X)$$
so using the identification $(l_g)_*$ is just left multiplication by $g$. 
Lastly note that $g X$ does not make sense a priori. What should it mean to multiply a derivation by a matrix? $gX$ is really just a shorthand notation for  $\Phi^{-1}(g\cdot\Phi(X))$ that is we identify, multiply and then identify back.
