I came across the following bit of a problem while I was preparing for my qualifying exams, and I am a bit curious about how you would go about proving that the $\eta$ given in the second sentence even existed.
Let $q:\mathbb{R}^2\rightarrow \mathbb{T}^2$ be the quotient map and consider the 1-form $\omega = 2\cos(\pi x)^2 dx + dy$ on $\mathbb{R}^2$. Then $\omega$ descends to a 1-form $\eta$ on $\mathbb{T}^2$ (i.e. there exists a 1-form $\eta$ on $\mathbb{T}^2$ such that $q^*\eta = \omega$).
You could try to compute something directly, but I don't really have an idea of how you would go about that, and I suspect that there is something more fundamental going on there. Specifically, I am trying to use the fact that, as $\mathbb{R}^2$ is the universal cover of $\mathbb{T}^2$, the map $q$ should act as a local diffeomorphism, but I am unsure how to proceed from there. Any explanation would be greatly appreciated; thank you so much!
The more fundamental thing that's going on here is the study of the map $p^* : \Omega^k(M/G)\to \Omega^k(M)$ where $M$ is a manifold, $k$ is some integer and $G$ is a Lie group acting nicely on $M$.
Precisely, $G$ acts freely and properly discontinuously on $M$ so that the quotient $M/G$ really is a manifold etc.
A thing one notices quite easily is that if $\omega\in \Omega^k(M/G)$, then $p^*\omega$ is $G$-invariant : $g^*p^*\omega = p^*\omega$. This is really easy to prove.
The interesting part is the converse : it's not always true, but sometimes it is.
In particular when $G$ is discrete the converse is true : any $G$-invariant $k$-form is of the form $p^*\omega$ for some $\omega$ (in fact you can show, with no discreteness hypothesis, that $p^*$ is injective, so $\omega$ is unique).
Here this is your situation with $\mathbb Z^2= G, M= \mathbb R^2$ and the translation action.
You should definitely try to solve this exercise for yourself. If you don't manage, I wrote a solution for $M=\mathbb R, G=\mathbb Z$ in this answer, and the general proof is essentially the same.