To your first/last question regarding the surjectivity of the covering map $p : \tilde X \to X$, you're right; in fact Hatcher explicitly states that a covering map need not be surjective. However, Hatcher also explains that the cardinality of the preimage $p^{-1}(x)$ for a given $x \in X$ is locally constant, i.e. this is the number of disjoint sheets in the preimage $p^{-1}(U)$ for a neighborhood $U$ of $x$. In particular, the definition of a covering space $p : \tilde X \to X$ does require that $p$ is surjective if $X$ is connected. Your example $X \sqcup Y$ is a valid example of a nonsurjective covering space, but can you think of why a connected space $W$ cannot be covered by both $X$ and $Y$ for appropriate choices of $X$ and $Y$? (Allen Hatcher pointed this user in the right direction.)
So I guess the answer to your last question is, perhaps it's not so much that there's an advantage to require or to not require that $p : \tilde X \to X$ is surjective. It's that there's an advantage to assuming that $X$ is connected (if it's not then you can just consider how the individual connected components of $X$ are covered by the different connected components of $\tilde X$). And if $X$ is connected, as discussed, $p : \tilde X \to X$ must be surjective.
I like to work out covering spaces by enumerating cosets. Recall that the fiber above the base point for a covering space associated with a subgroup $H\subset \pi_1(X)$ is in one-to-one correspondence with the set of cosets $\pi_1(X)/H$. Let $X=S^1\vee\mathbb{R}P^2$.
(c) is the most concrete without pictures, so I'll start here. Notice $H=\langle a,bab\rangle$ is a normal subgroup since $bab=bab^{-1}$. In the quotient $\pi_1(X)/H$, $a$ is trivial and $b$ remains, so it is $\mathbb{Z}/2\mathbb{Z}$. This is saying that $a$ lifts to a loop but $b$ does not. This must correspond to the cover $S^1\vee S^2\vee S^1$ (unfurling the $\mathbb{R}P^2$).
For (b) what you say is correct. I drew a Scheier graph for the subgroup $\langle b\rangle$, and it shows that it is the universal cover with the $b$ direction folded back. So, an $\mathbb{R}P^2$ in the middle, with the universal covers in each $a$ and $a^{-1}$ direction.
For (a), it might be easier to describe it as an $S^1$ attached to the $b$ direction of the universal cover. I imagine it as the universal cover rolled up in the $a$,$a^{-1}$ directions.
Coset enumeration is the same thing as taking the universal cover and folding it up based on the subgroup, which can be useful for group theory.
Perhaps this drawing of the universal cover will help:
This is obtained from the Cayley graph of the fundamental group generated by $a,b$ by replacing each $b$ edge with an $S^2$.
Best Answer
The covering space of $S^1 \vee S^1$ corresponding to the normal subgroup $N$ is the Cayley graph $\Gamma((\Bbb Z * \Bbb Z)/N)$ of $(\Bbb Z * \Bbb Z)/N$.
That this is a covering space is seen by noting that $\Gamma((\Bbb Z * \Bbb Z)/N)$ is homeomorphic to the orbit space $\Gamma(\Bbb Z* \Bbb Z)/N$, and the covering map $p : \Gamma(\Bbb Z* \Bbb Z)/N \to S^1 \vee S^1$ is given by mapping orbits of $N$-action on $\Gamma(\Bbb Z*\Bbb Z)$ to orbits of $\Bbb Z*\Bbb Z$-action on the graph.
Fact : If $X$ is a path connected, locally path connected and semilocally simply connected space and if a discrete group $G$ acts on $X$ freely and properly discontinuously, then there is a short exact sequence of fundamental groups $$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$$
Since action of $N$ on $\Gamma(\Bbb Z* \Bbb Z)$ is in fact free and properly discontinuous, $\pi_1(\Gamma((\Bbb Z*\Bbb Z)/N)) \cong N$. This proves that $\Gamma((\Bbb Z*\Bbb Z)/N)$ is indeed the covering space corresponding to $N$.