I found the following result on Wikipedia:
We can also find explicit generators for the de Rham cohomology of the torus directly using differential forms. Given a quotient manifold ${\displaystyle \pi :X\to X/G}$ and a differential form ${\displaystyle \omega \in \Omega ^{k}(X)}$ we can say that $\omega $ is ${\displaystyle G}$-invariant if given any diffeomorphism induced by ${\displaystyle G}$, ${\displaystyle \cdot g:X\to X}$ we have ${\displaystyle (\cdot g)^{*}(\omega )=\omega }$. In particular, the pullback of any form on ${\displaystyle X/G}$ is ${\displaystyle G}$-invariant. Also, the pullback is an injective morphism.
which states that the pullback $\pi^*: X/G \to X$ is injective. I was able to prove this but only in the case that the group $G$ is finite (it's somewhat of an "averaging" argument). But I can't see how to prove it in this general case (the "averaging" argument breaks down since a sum of infinite terms is not defined). I'd really appreciate some help on this!
When $G$ is compact, take a basis of left-invariant $1$-forms, wedge them to get a left-invariant volume form $dV(g)$, and given any $k$-form $\omega$, the averaged form $$\tilde\omega = \frac 1{\text{vol}(G)}\int_G L_g^*\omega\,dV(g)$$ is $G$-invariant. What's more, you can check that $\omega$ is closed/exact if and only if $\tilde\omega$ is closed/exact. So there's a unique invariant form in every cohomology class of $M$.
(Without compactness, the integral won't make sense, and nor will the volume.)