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
One image solution: here $g_1-1=2$, $g_2-1=6$, $n=3$. If $g_1$ is greater, just add holes in the middle. If $n$ is greater, just go on with the snake. As pointed out by John Hughes, you clearly need $g_1$ (and hence $g_2$) to be strictly positive.
Best Answer
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $\Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $\chi(\Sigma_{g,b}) = 2 - 2g - b$ and $\chi(S_{g,b}) = 2 - g - b$.
If $p : M \to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{\partial M} : \partial M \to \partial N$ of the same degree. So if $p : \Sigma_{g,b} \to \Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
\begin{align*} \chi(\Sigma_{g,b}) &= k\chi(\Sigma_{g',b'})\\ 2 - 2g - b &= k(2 - 2g' - b')\\ 2 - 2g - kb' &= k(2 - 2g' - b')\\ 2 - 2g &= k(2 - 2g')\\ \chi(\Sigma_g) &= k\chi(\Sigma_{g'}). \end{align*}
The converse is also true. That is, if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $\Sigma_{g,b} \to \Sigma_{g',b'}$. To see this, note that if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$, then there is a degree $k$ covering map $p : \Sigma_g \to \Sigma_{g'}$; see this answer. If $D \subset \Sigma_{g'}$ is the interior of a closed disc in $\Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $\Sigma_g$. So if $D_1, \dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $\Sigma_{g'}$, then $p^{-1}(\Sigma_{g'}\setminus(D_1\cup\dots\cup D_{b'})$ is $\Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $\Sigma_{g,b}$. Therefore the restriction of $p$ to $\Sigma_{g,b}$ is a degree $k$ covering $\Sigma_{g,b} \to \Sigma_{g',b'}$.
Likewise, $\Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g',b'}$ if and only if $\chi(\Sigma_g) = k\chi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings: