An $d$-sheeted unramified covering of compact Riemann surfaces induce a (monodromy) action on $d$ letters. Is the opposite true? Let $S_1, S$ be compact connected Riemann surfaces, $f : S_1 \rightarrow S$ be a meromorphic function of degree $d$ that branches over $B \subset S$. The unramified covering $f : S_1 \backslash f^{-1}(B) \rightarrow S \backslash B$ induces a (transitive) monodromy action of $\pi_1(S \backslash B, y)$ on the fibre $f^{-1}(y)$, which can be identified as an action on $d$ letters.
Is the inverse true? That is, fix $S$, $B$, $d$, and a transitive action of $\pi_1(S \backslash B)$ on $d$ letters. Now, can we build $S_1$ and $f$ that induce this action? Do we need extra conditions on the action for this to be possible?
 A: Yes, you can always do this, with one caveat.  Let $G\subset \pi_1(S\setminus B)$ be the stabilizer of a point in the monodromy action; then let $f:T\to S\setminus B$ be the covering space of $S\setminus B$ with fundamental group $G$.  Then $T$ will have all the necessary properties to be $S_1\setminus f^{-1}(B)$; the only question is whether $T$ can be compactified to a Riemann surface $S_1$ to which the map $f$ extends holomorphically.  But this question is local over each point of $B$; in particular, we can choose a small holomorphic disk $D_b$ around each $b\in B$ which does not contain any other points of $B$.  The inverse image $f^{-1}(D_b\setminus \{b\})\subset T$ is then a $d$-sheeted covering space of $D_b\setminus\{b\}$, and is thus itself a disjoint union of punctured disks, with $f$ looking locally like the map $z\mapsto z^n$ from $D\setminus\{0\}$ to itself, for $D\subset\mathbb{C}$ the unit disk.  We can thus fill in each of the punctures; doing this for each $b\in B$ yields a Riemann surface $S_1\supseteq T$ to which $f$ extends.  Furthermore, it is now easy to see that the extension of $f$ to $S_1$ is a proper map (since properness can be checked locally on the base), and hence $S_1$ is compact.
The one caveat is that the extension of $f$ to $S_1$ might not actually ramify over some points of $B$.  That is, the branch points of $f$ might be a proper subset of $B$ (for instance, there will never actually be any branch points if $d=1$).  So for the answer to be yes, what you need to mean by "$f$ branches over $B$" is that $f$ is unramified away from $B$, not that every point of $B$ actually is a ramification point.
