$(f\circ h)^* \omega = h^*f^*\omega$ - Legit now? Three pullback identities are required to prove on Guillemin and Pollack's Differential Topology on Page 164. I am not sure if I get it since it is the first time I play with them. 
I hope I got the first two right, but I was not able to proceed the third, because I don't see the commutativity of the pullback $df_x^*$ and the form $\omega$.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

$f^*(w_1 + w_2) = f^*w_1 + f^* w_2$
$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$
$\mathbf{(f\circ h)^* \omega = h^*f^*\omega}$
Following the definition, the RHS is:
\begin{eqnarray*}
d(f \circ h)_x^* \omega [(f \circ h)(x)]& =&\omega( d(f \circ h)_x(x))\\
& =&\omega( df_{h(x)} \circ dh_x)(x))\\
& =&\omega( df_{h(x)} dh_x(x))\\
& =&df_{h(x)}^*\omega( dh_x(x))\\
& =&f^*\omega( dh_x(x))\\
& =&h^*f^*\omega( x)\\
\end{eqnarray*}

Let $V_1, \dots, V_p \in T_x X$ be a collection of $p$ tangent vectors to $X$ at $x$. Then we compute that for any $x \in X$,
\begin{align}
((f \circ h)^\ast \omega)_x(V_1, \dots, V_p) & = \omega_{f(h(x))}(d(f \circ h)_x V_1, \dots, d(f \circ h)_x V_p) \\
 & = \omega_{f(h(x))}(df_{h(x)} dh_x V_1, \dots, df_{h(x)} dh_x V_p) \\
 & = (f^\ast \omega)_{h(x)} (dh_x V_1, \dots, dh_x V_p) \\
 & = (h^\ast f^\ast \omega)_x(V_1, \dots, V_p),
\end{align}
where we used the chain rule in going from the first line to the second line, and in the last two lines we just undid the definition of the pullback twice. Hence
$$(f \circ h)^\ast \omega = h^\ast f^\ast \omega.$$
