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
I think this is easiest to see using cellular homology. There is only one $2$-cell $U$, so $C_2 \cong \mathbb{Z}$. The boundary of this $2$-cell is the product of the commutators of the $2g$ $1$-cells: $[a_1, b_1] \cdots [a_g, b_g]$ where $[a_i, b_i] = {a_i}^{-1} {b_i}^{-1} a_i b_i$. For instance, if you traverse the boundary of the octagon you've drawn to represent a genus $2$ surface, you'll see that we traverse each edge twice, once in the positive direction, and once in the negative direction. See here for a reference. However, since the homology groups are abelian groups, these commutators are trivial, i.e.,
$$
\partial_2(U) = -a_1 - b_1 + a_1 + b_1 + \cdots + -a_g - b_g + a_g + b_g = 0 \, .
$$
Since $\partial_2$ maps the generator of $C_2$ to $0$, then $\partial_2 = 0$, hence $\ker \partial_2 = C_2 \cong \mathbb{Z}$.
Best Answer
Since Euler characteristic is multiplicative with respect to (finite sheeted) coverings, we see the orientable double cover must have twice the Euler characteristic. This already pins down the homeomorphism type of the cover.
To see it more explicitly, consider the usual embedding of an orientable surface into $\mathbb{R}^3$ (as a long chain-like shape). This embedding can be translated/rotated around so that the reflections in all 3 coordinate planes maps the surface to itself.
With this embedding, there is a $\mathbb{Z}/2$ action on $\mathbb{R}^3$ given by the antipodal map $x \mapsto -x$. This action preserves the embedded surface and acts freely on it so the quotient is a manifold. Since the antipodal map is orientation reversing, the quotient is nonorientable.