Chain rule of differential in smooth manifold I am having trouble understanding the chain rule in smooth manifolds (unfortunately that part of the book is an exercise). I see a resemblance to the chain rule in $\mathbb{R}^n$ but do not understand from the definition of the differential $dF_p(v_p)(f) = v_p(f\circ F)$ how to get there?
Ok, made a new try. Is this correct:
$$dF_p\circ v_p\circ f = v_p\circ f\circ F\\
d(G\circ F)_p\circ v_p\circ f = v_p\circ f\circ (G\circ F)=v_p\circ f\circ G\circ F\\
dG_{F(p)}\circ (dF_p\circ v_p)\circ f = (dF_p\circ v_p)\circ f\circ G=\\
=dF_p\circ v_p\circ (f\circ G) = v_p\circ (f\circ G)\circ F=v_p\circ f\circ G\circ F$$
I hope it is, because then pushforward makes sense all of a sudden. I guess it was all the parentheses that confused me.
 A: Before you try to prove it, you should try to understand why it should be true. In this case, if you have a smooth map $F:M\to N$, its differential $dF_p$ is the linearization of $F$ at $p$. Likewise, if $G:N\to O$, then $dG_q$ is the linearization of $G$ at $q\in N$. So $d(GF)_p$ should take the tangent space $T_pM$ to $T_{G(F(p))}O$ through $T_{F(p)}N$: it should be the linearization of $G$ acting on the image of the linearization of $F$, i.e., 
$$d(GF)_p = dG_{F(p)}\circ dF_p.$$
Now, as to proof details, your approach is good, but you should probably be a little bit more careful denoting what is acting on what: if $v_p\in T_pM$ and $f:O\to \mathbb{R}$, then
\begin{align*}
\big(d(G\circ F)_p(v_p)\big)(f) &= v_p( f\circ (G\circ F) )\\
&= v_p\bigg( (f\circ G)\circ F \bigg)\\
&= \big( dF_p(v_p)\big)\bigg( f\circ G\bigg) \\
&=  dG_{F(p)}\big(dF_p(v_p)\big)(f)\\
&= \bigg(\big(dG_{F(p)}\circ dF_p\big)(v_p)\bigg)(f)
\end{align*}
Since $v_p$ and $f$ were arbitrarily chosen, this establishes the result.
In words: if $v_p\in T_pM$ and $f:O\to\mathbb{R}$ are arbitrary, then the action of the vector resulting from differential of $GF$ applied to $v$ on $f$ is the action of $v$ on the composition of $f$ with $GF$. This is the same as the action of $v$ on the composition of $(fG)$ with $F$. That's the same as the action of $dF(v)$ on the function $fG$, which is the same as $dG(dF(v))$ acting on $f$.
You may also find it profitable to take one or two other approaches. One is to take local coordinate charts about $p$, $F(p)$, and $GF(p)$ and apply the fact from Euclidean calculus. Another is to take a curve $\gamma$ through $p$ and function $f$ and compute $\frac{d}{dt}(f\circ G\circ F\circ \gamma)(t)$.
