So, I've heard that if you have a finite degree covering of a compact connected manifold by another compact connected manifold of dimension $n$ (So $\pi :M \rightarrow N$) gives an injection on the deRham cohomology for $1\leq k\leq n-1$ given by $\pi^{*}$.
I am looking for a proof of this, but I am not sure where to find it. I skimmed through Lee's Introduction to Smooth Manifolds but did not find anything on the topic. I have heard that it has something to do with integration along fibres, but I am not sure.
Many Thanks!
You're on the right track. Since the fibers are finite in this case, saying "integration along the fibers" amounts to summing, or averaging, the preimages. In certain arenas, this is called the trace according to $\pi$.
Consider an open set $U\subset N$ that is evenly covered, and let $\phi_i\colon U\to U_i$ be (smooth) local inverses of $\pi$, $i=1,\dots,s$. Then for any closed $k$-form $\omega$ on $M$, we can consider the $k$-form on $N$ whose restriction to $U$ is given by $$\pi_*\omega\big|_U = \frac1s\sum_{i=1}^s \phi_i^*\omega.$$ Now I'll leave it to you to check that
(1) $\pi_*\omega$ is well-defined.
(2) If $\omega = \pi^*\psi$, then $\pi_*\omega = \psi$.
(3) $d(\pi_*\omega)=\pi_*(d\omega) = 0$, so $\pi_*\colon H^k(M)\to H^k(N)$.
More generally, whenever a compact Lie group $G$ (in this case a finite one) acts on $M$, you can show that $H^*(M) \cong H^*_G(M)$, where the latter is the cohomology of the complex of $G$-invariant differential forms. Taking $G$ to be the group of deck-transformations, we get $H^*_G(M)\cong H^*(N)$. (All cohomology here is deRham, so isomorphic to singular cohomology with $\Bbb R$-coefficients.)