A direct argument without the use of nets:
Let $\mathcal{C}$ be an open cover of $Y$. For each $p \in X$, choose an open set $p \in U \subseteq X$ such that $Y$ is trivial over $U$, and such that each lift of $U$ is contained in some element of $\mathcal{C}$. This is an open cover $\mathcal{D}$ of $X$, which has a finite subcover $\mathcal{D}'$ since $X$ is compact. The lift of $\mathcal{D}'$ to $Y$ is also a finite cover, as well as a cover that refines $\mathcal{C}$. Thus $\mathcal{C}$ must have a finite subcover. (The fact that $Y$ is a finite cover is used twice, first to make each $U$, second to lift $\mathcal{D}'$.)
Given any covering map $\Sigma_h\to\Sigma_g$ between two surfaces, is there some kind of a ``standard" covering $M\to \Sigma_g$, which factors through $\Sigma_h$?
In brief, the answer to this part of the question is 'You can take $M\to\Sigma_g$ to be regular, but beyond that, no.' This can already be extracted from Misha's comments above, but let me try to summarise.
First, note that there is a regular covering $M\to\Sigma_g$ that factors through $\Sigma_h\to\Sigma_g$. (Specifically, you can take $\pi_1M$ to be the intersection of all the conjugates of $\pi_1\Sigma_h$.) So, as in the earlier part of your question, you can take $\Sigma_h\to\Sigma_g$ to be regular.
You can now rephrase your question in terms of normal quotients of $\pi_1\Sigma_g$, and it becomes
Given any finite quotient $q:\pi_1\Sigma_g\to Q$, is there some kind of 'standard' finite quotient $p:\pi_1\Sigma_g\to P$ such that $q$ factors through $p$?
In particular, $P$ surjects $Q$. But any finite group can arise as $Q$ (see Misha's and algori's comments---the point is that $\pi_1\Sigma_g$ surjects a free group), so you are looking for a 'standard' family of finite groups that surjects every finite group. But there's no 'natural' definition of such a family.
Remark: Obviously, there are such families, such as $\{ Q\times\mathbb{Z}/2\}$ where $Q$ is an arbitrary finite group, but clearly this is not 'natural'.
Best Answer
Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$. We define $M_f = F \times I \,/\, f$ to be the mapping torus with monodromy $f$. That is, form the product $F \times I$ and identify the two ends via $f$. Note that $M_f$ is an $F$-bundle over the circle.
Now, the $(k+1)$-fold cover of $M_f$, obtained by "unwrapping the circle direction" is the mapping torus for $f^{k+1} = f$. Thus the $(k+1)$-fold cover is homeomorphic to $M_f$.
As a concrete example, the trefoil knot complement $X_T$ is a once-punctured torus bundle over the circle, with monodromy of order six. Thus $X_T$ seven-fold covers itself. (And also five-fold covers itself.) This trick works for any torus knot complement.