= Solution
A <topological surface> is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^2$.
For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the <real projective plane>, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.
In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore
$$
\boxed{S_2\text{ is not a topological surface}.}
$$
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