Thank you to Mariano Suárez-Alvarez. I think the cohomology ring will work, somthing like this:
We know that the cohomology ring of a genus $g$ surface has the property that any non-zero class in degree one has a class it can cup with to get a non-trivial element of $H^2$ (this could be from direct computation or by Poincare duality.) In the wedge sum, we notice that WLOG $X$ must have trivial $H^2$. Thus any two degree 1 class supported on $X$ cup to 0, and any class supported on $X$ cups to zero with a class supported on $Y$. Thus $X$ has trival $H^1$.
Another way to proceed, which is a bit different from polygonal presentations, is to think about connect sums. If $S$ and $T$ are surfaces, then we can form $S \# T$, a new surface, by cutting a disk out of each, and gluing together the resulting circle boundaries. For example, if $S$ and $T$ are both copies of $T^2$ (torus) then $S \# T$ is a copy of the genus two surface. Drawing some pictures will help here.
Let's use $S_{g,n,c}$ to denote the connected, compact surface obtained by taking the connect sum of a two-sphere with $g$ copies of $T^2$ (torus), $n$ copies of $D^2$ (disk), and $c$ copies of $P^2$ (projective plane, aka "cross-cap"). One standard notation is $N_c = S_{0,0,c}$ for the non-orientable surface of "genus" $c$.
Another way to obtain $N_c$ is as follows. Take the two-sphere $S_{0,0,0}$. Cut out $c$ disjoint closed disks to get $S_{0,c,0}$. Identify boundary component with the circle $S^1$. Finally quotient each boundary component by the antipodal map $x \mapsto -x$. This gives $N_c$. (More pictures!) Thus the orientable double-cover of $N_c$ is obtained by gluing two copies of $S_{0,c,0}$ along their boundaries, to get $S_{c-1,0,0}$. (Yet more pictures!)
Here are some closely related questions:
Showing the Sum of $n-1$ Tori is a Double Cover of the Sum of $n$ Copies of $\mathbb{RP}^2$
Covering space of a non-orientable surface
Actually, my answer above is a more abstract version of the second half of
https://math.stackexchange.com/a/279249/1307
Best Answer
The following relation is verified: $e(\Sigma_2)=|G|e(\Sigma_2/G)$ where $e(\Sigma_2)$ is the Euler characteristic. Since $e(\Sigma_2)=-2$, we deduce that $|G|$ divides $2$ and $G=1$ or $\mathbb{Z}/2$.
Euler characteristic of covering space of CW complex