How to compute the image of the fundamental group of a covering space of $S^1 \vee S^1$? In Allen Hather's Algebraic Topology, there is a section on the Covering Spaces of $S^1 \vee S^1$. The following picture is inside.

How do you rigorously compute $(1)$-$(9)$? 
If we take $E$ to be our covering space and $e_0$ to the the dark vertex, then we need to compute $\pi_1(E,e_0)$ in order to know the number of generators for our covering space. This can be computed using van Kampen's Theorem. Then let $p:(E,e_0) \to (S^1 \vee S^1, x_0)$ be our covering map. We wish to find the presentation of the image $p_{*}(\pi_1(E,e_0))$. What is the general method to compute $p_{*}(\pi_1(E,e_0))$? There is a very similar question here Covering spaces of $S^1 \vee S^1$, but the answers do not suffice.
 A: In general there is a Galois correspondence for covering spaces which tells you (under nice enough hypotheses on your space: just it being connected and the universal cover existing unless I'm forgetting something) that there is a bijective "inclusion reversing" correspondence between subgroups $H$ of $\pi_1(X)$ and covers $Y$ of the space $X$ where the subgroup $H$ is corresponds the fundamental group of the cover $Y$. Once you choose a basepoint upstairs in the universal cover you can obtain an action of $\pi_1(X, *)$ on $X^{univ}$ via deck-transformations of the cover, and all covers are obtained by taking the quotient $X^{univ}/H$ for some subgroup $H$. This is all in Hatcher's chapter on fundamental groups and covering space theory. 
The fundamental idea is that once basepoints are chosen paths and homotopies lift uniquely to the universal cover, thus a loop in the base space lifts to a unique path in the universal cover, which one can show is determined up to homotopy downstairs by its endpoints upstairs (rather: the correspondence it induces between points in the fiber upstairs that takes a point in the fiber over the base point $a_1$ to the point $a_2$ which is the unique endpoint of the unique lift of the loop starting at $a_1$), so the fundamental group of the base comes from a set of automorphisms of the fiber over the basepoint downstairs, which actually extend to covering automorphisms by the theory of Deck transformations.
In our case we can be more explicit though! The universal cover of $S^1 \vee S^1$ is just the infinite tree where every vertex has valence 4. Picking a basepoint vertex in this tree we can see the action of $\mathbb{Z} * \mathbb{Z}$ explicitly via travelling up and travelling to the right, which is obviously faithful by looking at where it sends our basepoint in the universal cover. Now simply take one of the subgroups written above, take the quotient of this tree by that subgroup by glueing, and verify that what you get is the shape Hatcher draws. To be helpful I'll do the first one, the rest are up to you!
For (1) take the subgroup $<a>$ coming from shifting our basepoint to the left or right. Then quotienting by $<a>$ identifies all of the vertices and edges of the tree directly to the left or right of the basepoint and identifies all of the other vertices and edges that can be obtained by taking a given path from a horizontal translate of the basepoint. It's hard to explain in words so I drew a little picture below, the different colors denote different vertices that are identified and the arrows denote the paths which become the nontrivial loop in the quotient. 
You can verify that the quotient is given in the picture below:
Similarly we mod out by the action of $H_1 = <b^2, a>$ and we get the doodle below:

Finally we need to understand how to take the quotient by $H = <b^2, a, bab^{-1}>$. Lets think about what further identifications we need to make on the vertices of the tree. First note all of the unidentified vertices are in bijection with cosets $x \cdot H_1$ where $x = (\dots) \cdot b$ via the action of $\pi_1(X) = G$ on the fiber over the point where the two circles meet in $S^1 \vee S^1$. Now lets think about this one algebraically, what elements are in $H$ and thus what representatives are there for cosets of $G/H$? Well $bab^{-1}, b^2 \in H$ thus $bab \in H$ as $bH = b^{-1}H$. Further this works for any power of $a$ as $(bab^{-1})^n = ba^nb^{-1}$, so we also have that $ba^nb \in H$ this shows us that any representative for a nontrivial coset can be put in the form $a^nb$ for some $n \in \mathbb{N}$, but as $bab^{-1} \sim a$ modulo $H$ we have that $ab = ba = b$ modulo $H$. So all of the remaining "tree" part of our space gets identified to a loop one unit up from the basepoint. All this together gives us:

Where the "top" version of $a$ is actually representing the loop coming from the vertex $b \cdot *$ where $*$ is the basepoint of $X^{univ}$. I hope this helped. In general these things are very topologically intuitive but difficult to explain systematically without just entirely reducing it to algebra, I tried to do a little bit of both here to show you multiple ways of thinking about this type of problem. Computing the correspondence between a covering space and a subgroup will simply consist of either carefully computing the quotient (if you know the group but not the cover) in the manner above or (if you know the cover but not the group) arguing rigorously about which loops in the base lift to loops upstairs. Feel free to ask more questions in the comments if you like.
P.S.: Note that the fact that this cover has as many symmetries as it has points in a fiber means that the subgroup we found was actually normal by the Galois correspondence. This is obvious for other reasons but this gives a nice topological proof.
