Recall the theorem which says that the group of deck transformation of the universal cover is isomorphic to the fundamental group of your original space. If you chase through the proof with the example in Hatcher you might get more of an intuition.
Geometrically, you could reason as follows. $\mathbb{R}P^2$ is obtained by taking the quotient of $S^2$ under the antipodal map. So $\mathbb{R}P^2\vee \mathbb{R}P^2$ is the wedge union of two spheres, with each sphere then quotiented by the antipodal map.
Choose the point of identification as a base point. Then a path $a$ which goes halfway round the equator of one of the spheres goes all the way around one copy of $\mathbb{R}P^2$, since antipodal points are identified. Thus it is a loop. Similarly there's a loop $b$ which goes around the other copy of $\mathbb{R}P^2$. These naturally correspond to the $a$ and $b$ in Hatcher's picture.
In particular the deck transformation $ab$ corresponds to the loop going once around one copy of $\mathbb{R}P^2$ and then once around the other copy.
Warning: these kind of geometrical arguments are not always easy. If you aren't careful you can give yourself wrong ideas about a space. I'd recommend trying to get a good feel for how the techniques of covering spaces work abstractly. Once you've derived a result you can then use it to inform your geometrical intuition.
Good luck with learning about covering spaces. They are a bit confusing at first, but they're really cool when you get to know them!
I believe I have figured out the answer.
To give a "concrete description" of all the $n$-sheeted covering space up to isomorphism, we need to know all the $m$-sheeted covering spaces of $X$ and $Y$ for $m\leq n$. This includes $X$ and $Y$ themselves. Specifically, an $n$-sheeted covering space of $X\vee Y$ is a connected "graph" with the following properties:
- There are $n$ edges
- Given a vertex of our graph of valence $m$, that vertex is an $m$-sheeted connected covering space of either $X$ or $Y$.
- No edge connects two covering spaces of $X$ or two covering spaces of $Y$.
Specifically, at each vertex, the edges attach at points belonging to the preimage of the basepoint $x_0\in X\vee Y$. We can pick any of these edges to be a basepoint of the covering space. Namely, once this graph is constructed, retract each of the lines connecting the vertices that make up the "edges" to points. For example, a graph with two vertices would become the wedge sum of two spaces.
A concrete example: consider the space $B=\mathbb RP^2\vee T^2$ (where $T^2=S^1\times S^1$ is the torus). $B$ has $7$ isomorphism classes of connected two-sheeted covering spaces. This can be seen from the fact that:
- $\mathbb RP^2$ has one two-sheeted covering space, namely, $S^2$.
- $T^2$ has three two-sheeted covering spaces, specifically, those corresponding to the subgroups $\langle a^2,b\rangle$, $\langle a,b^2\rangle$, and $\langle a^2,ba^{-1}\rangle$ of $\pi_1(T^2)=\mathbb Z^2=\langle a,b\mid [a,b]\rangle$ (these can be seen as the kernels of epimorphisms $\mathbb Z^2\to\mathbb Z_2$). Let's call these covering spaces $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$,
There are two possible connected graphs with $2$ edges, namely, that with three vertices obtained by connecting two line-segments end-to-end, and that with two vertices, each vertex of valence 2 with both edges leading to the other vertex.
In the former case, if middle vertex (of valence 2) is a two-sheeted covering space of $\mathbb RP^2$, of which there is only one choice, then the outermost vertices are both $T^2$. On the other hand, there are three possibilities for the middle vertex if it is a covering space of $T^2$, namely, $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$. The outermost vertices must be $\mathbb RP^2$. Hence, there are $4$ possible $2$-sheeted covering spaces of $B$ with three "vertices."
In the latter case, we have two vertices each of valence two. For the vertex that covers $T^2$, we have three choices, and for the vertex that covers $\mathbb RP^2$. we have only one choice, the sphere. In this way, we obtain $3$ more covering spaces of $B$.
I have drawn each of these graphs and labelled the vertices accordingly below. Red vertices correspond to covering spaces of the torus, while blue vertices correspond to covering spaces of $\mathbb RP^2$.
Best Answer
The construction is inductive. I will use the principle that "et cetera is permitted" which I learned in freshman number theory.
Let $p \in X$, $q \in Y$ be the points which are identified to each other in $X \vee Y$. Start with a copy of the universal cover $\widetilde X$. To each lift $\tilde p$ of $p$, attach a copy of $\widetilde Y$ by identifying that $\tilde p$ to some $\tilde q$. To each unidentified $\tilde q$, attach a copy of $\widetilde X$ by identifying that $\tilde q$ to some $\tilde p$. To each unidentified $\tilde p$, attach a copy of $\widetilde Y$… To each unidentified $\tilde q$, attach a copy of $\widetilde X$… Et cetera, et cetera.
I wrote up this construction as a solution to a different problem but have been unable to locate it. It seems the question has been deleted, or my ability to search is inadequate.