I believe that this should hold true:
Result. Let $\pi: M \to N$ be a Riemannian covering ($M$,$N$ closed). Then $\Delta_M \circ\pi^* = \pi^* \circ \Delta_N$.
This should be true, but I could not find it properly stated or proved.
I thought about using the fact that a covering is a local isometry, and so on evenly covered open sets, since laplacian commutes with isometries, I have the result. But now I struggle to see how I should patch things up to get a global result.
I'm reading a paper on spectral geometry and I want to relate eigenspaces of $\Delta_M$ with the ones of $\Delta_N$. I would appreciate some hints.
I hope I'm not misinterpreting your issue, but the way I read it seems like you've done the hard part already.
When you say you can prove the result for evenly covered open sets, you must mean that you have showed that if $U$ is an evenly covered subset of $N$ and $V$ is a component of its preimage in $M$, then for $f$ a (differentiable enough) function on $U$, $p^{\ast} \Delta_N f = \Delta_M p^{\ast} f$, where $p$ is the isometry between $U$ and $V$. But this should give you the result globally, since $V$ was an arbitrary component and for $x \in V$, $p^{\ast} g (x)= \pi^{\ast} g(x)$ for any function $g$ on $U$. In other words, $\pi$ does the patching for you.