I am looking for a proof verification on the following problem:
Let $M : = S^1 \times S^1$ with coordinate chart (inverse) $\phi : (0,2 \pi) \times (0, 2\pi) \rightarrow M$ given by $\phi(x , y) : = (e^{ix} , e^{iy})$. Show that $dx \wedge dy$ extends to a global 2-form that represents a nontrivial class in $H_{\text{dR}}^2(M)$.
In order to extend $dx \wedge dy$, let $\psi: (\pi, 3 \pi) \times (\pi, 3 \pi) \rightarrow M$ be given by $\psi(u , v) = (e^{iu} , e^{iv})$. It is not hard to show that $M =\operatorname{Im} \phi \cup \operatorname{Im} \psi$, that the transition maps between $(x,y)$ and $(u,v)$ are
\begin{eqnarray*}
x = u - \pi \quad \quad \quad \quad u = x + \pi \\
y = v - \pi \quad \quad \quad \quad v = y + \pi\\
\end{eqnarray*}
and $dx \wedge dy = du \wedge dv$ on $\operatorname{Im} \phi \cap \operatorname{Im} \psi$. If we define
$$
\omega : =\begin{cases}
dx \wedge dy & \text{ on } \operatorname{Im} \phi\\
du \wedge dv & \text{ on } \operatorname{Im} \psi\\
\end{cases}
$$
then $\omega$ is well-defined because $dx \wedge dy = du \wedge dv$ on $\operatorname{Im} \phi \cap \operatorname{Im} \psi$, smooth, and extends $dx \wedge dy$. Moreover, $\omega$ is clearly closed, but not exact because
$$
\int_{M} \omega = \int_{0}^{2\pi} \int_{0}^{2\pi} dx\,dy = 4 \pi^2;
$$
hence $\omega$ represents a nontrivial class in $H_{\mathrm{dR}}^2(M)$.
Obviously some details of the calculation are omitted, but is this correct or is there anything worth adding?
One problem is that $M \ne \text{Im} \phi \cup \text{Im} \psi$, for example the point $(e^{i\pi},e^{i0}) = (-1,+1) \in M$ is not in either of those images, nor is the point $(e^{i0},e^{i\pi})=(+1,-1)$. However, it looks like those are the only two points missed out of all points in $M$.
I think you can patch that up, and once that is done, the rest of your answer should hold up.