I was quite puzzled by the request of classifying all the $4$-covers of the connected sum of $5$ copies of $\Bbb R P^2$. For oriented covering space, the answer is well known: it's enough to consider the $2$-cover of the orientable cover. But I was quite surprised I wasn't able to find the non-orientable covering (or rule it out).
In general, what can we say about non-orientable finite coverings spaces of non-orientable surfaces?
If $M$ is a connected manifold, let $nM$ denote the connected sum of $n$ copies of $M$.
First note that if $X \to Y$ is a $k$-sheeted covering, then $\chi(X) = k\chi(Y)$. So if $X$ is a four-sheeted covering of $5\mathbb{RP}^2$, $\chi(X) = 4\chi(5\mathbb{RP}^2)$. Now recall that $\chi(n\mathbb{RP}^2) = n - 2$, so $\chi(X) = 4(5 - 2) = 12$. If $X$ is non-orientable and connected, then $X = n\mathbb{RP}^2$ for some $n$. But $\chi(X) = 12$ and $\chi(X) = \chi(n\mathbb{RP}^2) = n - 2$, therefore $n = 14$. That is, the only connected non-orientable surface which can be a four-sheeted cover of $5\mathbb{RP}^2$ is $14\mathbb{RP}^2$. Note, we still need to check whether $14\mathbb{RP}^2$ does cover $5\mathbb{RP}^2$ or not.
Here's a useful trick: if $\pi: X' \to X$ is a $k$-sheeted covering, then there is a $k$-sheeted covering $p: X'\# kY \to X\# Y$.
To see this, let $U$ be an evenly covered open disc in $X$, then $X'\setminus\pi^{-1}(U)$ is $X'$ with $k$ disjoint open discs removed. Let $V$ be an open disc in $Y$. Gluing a copy of $Y\setminus V$ to each boundary sphere of $X'\setminus\pi^{-1}(U)$, and a copy of $Y\setminus V$ to the boundary sphere of $X\setminus U$, the $k$-sheeted covering $X'\setminus\pi^{-1}(U) \to X\setminus U$ extends to a $k$-sheeted covering $X'\# kY \to X\# Y$.
Returning to the problem at hand, recall that $3\mathbb{RP}^2 = T^2\#\mathbb{RP}^2$, so $5\mathbb{RP}^2 = T^2\# 3\mathbb{RP}^2$. There is a four-sheeted covering $T^2 \to T^2$ (for example, $(z, w) \mapsto (z^4, w)$), so by the above construction, there is a four-sheeted covering $T^2\# 12\mathbb{RP}^2 \to T^2\# 3\mathbb{RP}^2$, i.e. $14\mathbb{RP}^2 \to 5\mathbb{RP}^2$.
If the deck transformations of the four-sheeted covering map $T^2 \to T^2$ are generated by a $90$ degree rotation (which is the case for the example I gave), you can visualise the construction of the four-sheeted covering of $T^2\# 3\mathbb{RP}^2$ as follows:
$$3\mathbb{RP}^2 \\ \# \\ 3\mathbb{RP}^2\ \#\ \ T^2\ \#3\ \mathbb{RP}^2 \\ \# \\ 3\mathbb{RP}^2$$
These arguments can be used to show the following:
Let $X$ and $Y$ be two closed, connected surfaces, either both orientable or both non-orientable. Then there is a $k$-sheeted covering $X \to Y$ if and only if $\chi(X) = k\chi(Y)$.
As every covering of an orientable space is orientable, the only case left to consider is orientable coverings of non-orientable surfaces.
Let $X$ and $Y$ be two closed, connected surfaces, with $X$ orientable and $Y$ non-orientable. Then there is a $k$-sheeted covering $X \to Y$ if and only if $\chi(X) = k\chi(Y)$ and $k$ is even.
The reason we need $k$ to be even in this statement is that every orientable covering of a non-orientable manifold factors through the orientation double cover.
There certainly exist some: here, for example, is a non-orientable map of 110 pentagons, which is a 2-fold cover of this map. There seem to be plenty of other examples of non-orientable covering maps.
