These results are illuminated by the notion of covering morphism of groupoids and the result that:
If $G$ is a groupoid, then there is an equivalence of categories between covering morphisms of $G$ and actions of $G$ on sets.
If $X$ is a suitably nice space, then the fundamental groupoid functor gives an equivalence of categories between covering maps of $X$ and covering groupoids of $\pi_1(X)$.
Note that because this models a map of space by a morphism of groupoids, this model is somewhat nearer to intuition, and is easier to apply, than the corresponding action on sets. These ideas are applied in a recent paper
Jeremy Brazas, Semicoverings: a generalization of covering space theory, Homology, Homotopy and Applications, 14(1), (2012) 33-63. (downloadable)
This question is related to, roughly speaking, how "symmetric" your covering space is. Specifically, you know a subgroup will be normal iff the quotient is actually a group. In covering spaces, this corresponds to being having homeomorphisms of the covering space that leave the covered space unaffected via the covering map. This type of question gives 2 ways to argue the solution, since there is an order-reversing bijection between connected covering spaces and subgroups of the fundamental group.
Topologically: For this specific example, I'm going to choose the basepoint of $S^1\vee S^1$ to be the wedge point, $b_0$. Then for each covering space $p:E\rightarrow S^1\vee S^1$, we can choose our favourite basepoint $e_0\in p^{-1}(b_0)$, and determine if for each $e\in p^{-1}(b_0)$ there is a homeomorphism $h_e:E\rightarrow E$, subject to $h_e(e_0)=e$ and $p\circ h_e=p$.
Algebraically: We end up with $\pi_1(E,e_0)\hookrightarrow\pi_1(S^1\vee S^1, b_0)$, and identifying $\pi_1(E,e_0)$ with its image, we get $\pi_1(S^1\vee S^1, b_0)/\pi_1(E, e_0)$ is a group iff $\pi_1(E, e_0)$ is normal in $\pi_1(S^1\vee S^1, b_0)$ iff $p:E\rightarrow B$ is a normal covering space.
For the first covering space, I choose $e_0$ to be the center of the picture drawn, and simply by moving the picture left/right/up/down, we can see a homeomorphism of the space taking the center crossing to any other crossing we choose, while still preserving the space. However, this case is trivial on the algebra side, since the covering space is simply connected, so has trivial fundamental group, so corresponds to a normal (trivial) subgroup of $\pi_1(S^1\vee S^1,b_0)$.
The second covering space, however, is not normal, since almost any homeomorphism of our required type cannot preserve the single loop. On the algebra side, this says that $\langle a \rangle<\langle a,b\rangle$ is not a normal subgroup, which is true since $bab^{-1}\neq a^n$ for any $n$.
Now, the third covering space has the desired symmetry, since there are only two elements in the preimage of our basepoint, and the homeomorphism taking one to the other is a simple mirroring (switching the left and right sides) of our picture across its center. On the algebra side, this says that a subgroup of $\pi_1(S^1\vee S^1,b_0)$ of index 2 is normal, which is always true.
The final covering space is also normal, since we can send any basepoint to any other via a translation of our picture. On the algebra side, this covering space (labeling all lower half circles $a$ moving left to right, and all upper-half circles $b$ moving right to left) corresponds to the subgroup $\langle a^n(ab)a^{-n} \colon n\in\Bbb Z\rangle<\langle a,b\rangle$, which is normal because $a(a^n(ab)a^{-n})a^{-1}=a^{n+1}(ab)a^{-(n+1)}$ and $$b(a^n(ab)a^{-n})b^{-1}=(a^{-1}(ab)a)(a^{n-1}(ab)a^{-(n-1)})(a^{-1}(ab)a)^{-1},$$ both of which are in our subgroup.
Best Answer
I think the most natural way of doing this is to define first a covering morphism $p: Q \to G$ of groupoids; the definition is essentially unique path lifting: more specifically, we require that if $x \in Ob(Q)$ and $g \in G$ has source $px$, then there is a unique $h \in Q$ with source $x$ such that $p(h)=g$. There are several main results:
A covering map of spaces induces a covering morphism of fundamental groupoids.
The category of covering morphisms of a groupoid $G$ is equivalent to the category of actions of $G$ on sets.
If $X$ is a space, and $q: Q \to \pi_1 X$ is a covering morphism of groupoids, then under certain local conditions on $X$ there is a topology on $Ob(Q)$ such that the map $Ob(q)$ becomes a covering map and ...(I'll leave you to fill in the rest!).
This treatment was in the 1968, 1988 editions of my book which is now available as Topology and Groupoids. The nice point is that a map of spaces is well modelled by a morphism of groupoids; this is convenient for questions on liftings. Also, under the right local conditions, you get an equivalence of categories from covering spaces of $X$ to covering morphisms of $\pi_1 X$.
This is also an introduction to the useful notion of fibration of groupoids.
I think I should mention that the book also has a treatment of orbit spaces and orbit groupoids, which you won't find elsewhere.
Later: A related use of covering morphisms of groupoids is in the paper:
J. Brazas, "Semicoverings: a generalisation of covering space theory", Homology, Homotopy and Applications", 14 (2012) 33-63.