Let $M$ be a topological manifold such that $M$ is homeomorphic to the interior of a compact manifold with boundary. Let $f \colon M' \to M$ be a finitely-sheeted covering.
Is then also $M'$ homeomorphic to the interior of a compact manifold with boundary?
First, the terminology: A topological manifold $M$ is called tame if it is homeomorphic to the interior of a compact topological manifold with boundary. Here is one nontrivial fact that you need to know: If $N$ is a topological manifold with boundary then the inclusion map $int(N)\to N$ is a homotopy-equivalence. This is a corollary of the existence of a collar around the boundary of $N$, see my answer here. In particular, given $x\in int(N)$, the induced map $\pi_1(int(N),x)\to \pi_1(N,x)$ is an isomorphism. (One can in principle prove it with bare hands, without using Brown's theorem, but why bother...)
Once you know this, the rest is just the basic covering theory which you can find either in Hatcher or in Massey.
Suppose that $M$ is a connected tame manifold (I will leave it to you to deal with the case of non-connected manifolds), $N$ is a compact manifold with boundary, $f: M\to int(N)$ is a homeomorphism. Let $p: M'\to M$ be a covering map (finite or not is irrelevant at this point). Then $p$ is determined by a certain subgroup $H\le \pi_1(M,y)$. Using the isomorphism (he we need Brown's theorem) $f_*: \pi_1(M,y)\to \pi_1(int(N), x)\cong \pi_1(N, x)$, $x=f(y)$, we obtain a subgroup $G=f_*(H)\le \pi_1(N, x)$. Now, let $q: N'\to N$ be the covering determined by $G$. Clearly, $M'$ is homeomorphic to the interior of $N'$ (via a lift of $f$). Restricting to the case of finite coverings, $H$ is a finite index subgroup of $\pi_1(M,y)$, hence, $G$ is a finite index subgroup of $\pi_1(N,x)$, hence, $q: N'\to N$ is a finite covering, hence, $N'$ is a compact manifold with boundary (I trust you know how to prove this). Thus, $M'$ is tame. qed