I am in trouble with the proof of this theorem:
Let $\Lambda^m(T^*M)$ be trivial: then there exist a nowhere vanishing $\omega$ $\in$ $\Omega^m(M)$
In particular at the beginning of the proof a generic atlas is considered:
Let ${(U_{\alpha},\Psi_{\alpha})}$ be any atlas (not necessarily oriented):
$\Rightarrow$ $\forall \alpha$ $\Psi_{\alpha}^*(dx_1\wedge...\wedge dx_m)= f_\alpha \cdot \omega$
where $f_\alpha$ : $U_{\alpha}$ $\longrightarrow$ $\mathbf{R}$ is non-vanishing.
What I don't understand is this last equality and also why $\Psi_{\alpha}^*(dx_1 \wedge ... \wedge dx_m)$ is non vanishing.
Presumably, $m=\dim M$. The existence of such a non-vanishing form is immediate: by assumption, there is a bundle isomorphism $$ F\colon \Lambda^m(T^*M) \overset{\sim}{\to} M\times \Bbb R. $$ Consider the constant section $s$ of $M\times \Bbb R$ given by $s(p) = (p,1)$. It is a nowhere vanishing section. Now, $\omega = F^* s$ is a nowhere vanishing section of $\Lambda^m(T^*M)$.
Once such a non vanishing section $\omega$ have been proven to exist, for any (local) differential $m$-form $\sigma$, there exists a function $f$ such that $\sigma = f \omega$. This is because $\Lambda^m(T^*M)$ is a rank one vector bundle, and therefore, $\omega(p)$ is a basis of the one dimensional vector space $\Lambda^m(T^*M)_p$, for all $p$. You can apply this to $\varphi^*(dx^1\wedge \cdots \wedge dx^m)$ for any chart $\varphi$.