I'll give here another answer which I think is correct but I am not 100% sure. We will rely heavily on the first theorem at page 11 of Hatcher's Algebraic Topology, which basically allows us to do 2 things:
- We can kill any contractible line without changing homotopy type
- Instead of identifying two points we can just attach a $1$-cell at these two points.
Our first observation is that after identifying the edges, non of the vertices of the large square has been identifyied with eachother therefore we can collapse each edge of the edges $AB,BC,CD,DA$ to a point. The result space is Space $1$, where red points means identified (see the picture).
Now, we get the homotopy equivalent Space $2$ by replacing identifications by $1$-cells.
At last, we pull the outside point to the center, to get Space $3$, where the red lines are to be thought above the shape.
Now for the tricky part: The disc, without the red lines, deformation retracts to the wedge of $4$ circles. To picture this, imagine a strong gravity well at the center of the disk pulling everything radially at the center of the disc. Evertyhing either falls to the boundaries of the 4 existing holes or at the center of the disc, including the attaching points, therefore the red lines now became circles.
Therefore we end up wit the wedge of 8 circles and the fundamental group is $F_8$.
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
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.