Consider the following degree $4$ non-normal covering space of $S_1\lor S^1$ I drew:
Here, $a$ and $b$ denote the edges which map onto the first and second circle in $S^1\lor S^1$ respectively. I believe that the deck group for this covering space is $\mathbb{Z}/2\mathbb{Z}$, with the only nontrivial deck transformation being a "$180^\circ$ rotation". Now, there is a theorem that $\operatorname{Deck}(p)\cong N(H)/H$ where $p:E\to S^1\lor S^1$ is the covering map, $H=p_*(\pi_1(E,e))$, and $N(H)$ is the normalizer of $H$ in $\pi_1(S^1\lor S^1,x)\cong F_2$. So I wanted to compute this with this example. However, I am having trouble determining exactly what $H$ is, and hence also what $N(H)$ is. How can I compute, for example, a minimal generating set of $H$ and then compute $N(H)$?
Edit: By contracting some edges, I was able to determine that the covering space $E$ in the picture is homotopy equivalent to $S^1\lor S^1\lor S^1\lor S^1\lor S^1$. By taking the five generators of this space and pulling them back through the homotopy equivalence, I was able to get that a generating set of $H$ is $H=\langle aba^{-1},a^2b,b^2,ba^2,baba^{-1}b^{-1}\rangle$. The question now is whether this is a minimal generating set, and how to compute $N(H)$ using it.
Edit 2: It seems that $bHb^{-1}=H$, so $b\in N(H)$. I therefore conjecture that $N(H)=\langle aba^{-1},b,a^2\rangle$. It follows from my observation that $N(H)\supseteq \langle aba^{-1},b,a^2\rangle$ but I am not sure how to show equality. If equality does hold, then we see $N(H)/H\cong\mathbb{Z}/2\mathbb{Z}$, agreeing with my observation that the deck group seems to be $\mathbb{Z}/2\mathbb{Z}$.
First of all, subgroups of free groups are free (see Nielsen-Schreier theorem). Moreover, if $H \subset F_n$ is a subgroup of the free group $F_n$ with $n$ generator and $H$ has finite index $k$ in $F$, then $H$ is isomorphic to $F_{k(n - 1) + 1}$ so in you case, $H$ is isomorphic to $F_{4(2 - 1) + 1} = F_5$ and, indeed, all the covering spaces of $S^1 \lor S^1$ of degree $4$ are homotopic to $S^1 \lor S^1 \lor S^1 \lor S^1 \lor S^1$.
Therefore, as you found $5$ generators to $H$, they are free and hence the minimal number of generators in $H$. Let $G = \left<H,b\right> = \left<aba^{-1},b,a^2\right>$ which is a subgroup of $N(H)$ containing $H$ as you noticed.
Using the previous formula which says that if $K \subset F_m$ has finite index $k$, then $K = F_{k(m - 1) + 1}$, we deduce in particular that $K$ is generated by more elements than $F_m$ ($k(m - 1) + 1 \geqslant m$ when $k,m \geqslant 1$). As $F_3 = G \subset N(H) \subset F_2$ and we always have finite indices, then $N(H)$ either has $2$ or $3$ generators. If it has $2$, $N(H) = F_2$ else $N(H) = G$.
But $H$ is not normal in $G$ so $N(H) = G$.