 Solution

A topological surface is a Hausdorff second-countable space locally homeomorphic to ${\mathbb{R}}^2$. Pairing opposite sides of the hexagon gives neighborhoods modelled on discs, including at edge and vertex classes; compactness follows from the compact polygon quotient. Cutting and rearranging the polygon gives the usual opposite-side square, hence a torus.

The full double cone is not a surface at the origin: deleting the vertex from a small neighborhood leaves two components, unlike a punctured disc. The upper cone with $z>0$ excludes that vertex and has ordinary annular charts, so it is a surface.