Wrong Answer
Given a finite connected "nice" graph, $G$, you can take all triples $\{a,b,c\}$ of nodes with $\{a,b\}$,$\{b,c\}$, and $\{a,c\}$ edges in the graph.
Take these as $2$-simplexes, and stitch them together in the obvious way.
The fact that $G$ is nice means that each edge must be on exactly two triangles. The fact that $G$ is nice also means that the interior of the union of the triangles that contain node $a$ will be homeomorphic to an open ball in $\mathbb R^2$.
So this all shows that stitching these together will yield $G$ as a triangulation of a compact $2$-manifold.
There is at least one "degenerate" case for which this is not true - the single-edge graph with two nodes. Depends on whether you consider a single node graph to be a cycle...