Elaborating on some comments:
Let $S$ be the Riemann sphere. Consider triples $(a,g,b)$ where $a\in S$, $b\in S$, and $g$ is the germ at $a$ of an analytic map from a neighborhood of $a$ to $S$ that takes $a$ to $b$. Make the set of all such triples into a one-dimensional complex manifold as follows.
(It is not a manifold in the strictest sense, because it has uncountably many connected components, but we will be interested in just one of its components.)
A chart is given by any pair $(U,f)$ where $U\subset S$ is open and $f:U\to S$ is analytic. Use the obvious bijection between $U$ and set of all points $(a,g,f(a))$ where $a\in U$ and $g$ is the germ of $f$ at $a$.
So given the germ, at some point $a\in S$, of a meromorphic function, you get a connected one-dimensional complex manifold, whose points are all the germs that can be obtained by analytic continuation from that one.
Usually this is not compact, but if what you start with a pair $(U,f)$ that corresponds to an open subset of $\Sigma$ and the restriction of a global map $\Sigma\to S$ then it will be, if not $\Sigma$, then some compact Riemann surface that has $\Sigma $ as a branched cover.