This exercise from Hatcher's book is just an application of group theory:
Exercise (Part 1). For a path-connected, locally path-connected, and semilocally simply-connected space $X$, call a path-connected covering space $E\to X$ abelian if it is normal and has abelian deck transformation group. Show that $X$ has an abelian covering space that a covering space of every other abelian covering space of $X$, and that such a 'universal' abelian covering space is unique up to isomorphism.
Proof. To prove existence, take $N=\pi_1(X)'=[\pi_1(X),\pi_1(X)]$. Then there exists a covering space $p':X'\to X$ uniquely determined by $N$. It's normal since $X$ is path-connected and $N\trianglelefteq \pi_1(X)$, and it's abelian since $G/G'$ is abelian for any group $G$. Furthermore every subgroup $H\trianglelefteq G$ such that $G/H$ is abelian contains the commutator $G'$, so if $p:E\to X$ is an abelian covering space, then
$$H=\im p'_{*}\leq \im p_{*}=\pi_1(E)$$
so $p'$ lifts to a map $\tilde{p}':X'\to E$ which, by the lemma 1234567 here is a covering map. We're now done with the existence. To prove uniqueness, let's suppose $Y$ is such a covering space. Then since $X'$ is abelian, $Y$ is a covering space of $X$, so that $\pi_1(Y)\leq N$. Furthermore $Y$ is abelian, so $N\leq \pi_1(Y)$. Thus $N=\pi_1(Y)$ and by uniqueness of $X'$ as the covering space associated by the commutator, $X'\cong Y$.
Exercise (Part 2). Describe this covering space explicitly for $X=S^1\vee S^1$ and $X=S^1\vee S^1\vee S^1$.
Solution. When $X=S^1\vee S^1$, take $E=\bR\times \bZ\cup \bZ\times \bR$ the integer grid of $\bR^2$ and let $\bZ\times \bZ$ act on $E$ by $(m,n)(x,y)=(x+m,y+n)$. Furthermore, each $(x,y)\in E$ has a neighborhood $U$ such that $(m,n)U\cap U\neq \varnothing$ only when $(m,n)=(0,0)$, so this action is a covering space action. So the quotient map $p:E\to E/\bZ\times \bZ$ is a normal covering map and $\bZ\times \bZ$ is the group of deck transformations of $p$. But $E/\bZ\times \bZ\cong X$. This proves $E$ is an abelian cover of $X$. The generating loops $a,b$ of $X$ are lifted to $E$ at each point $(i,j)$ of the fiber by $\tilde{a}:(i,j)\leadsto (i+1,j)$ and $\tilde{b}: (i,j)\leadsto (i,j+1)$. By a quick use of Van-Kampen's Theorem each loop in $E$ can be seen as a product of loops in Tic-Tac-Toe-shaped open sets, which are homotopic to loops in unit squares. But each of these is just of the form $(\tilde{\alpha}\tilde{\beta}\tilde{\overline{\alpha}}\tilde{\overline{\beta}})^n$ where $\tilde{-}$ means a lift of $-$ at some point, and from this follows that the elements of $p_*\pi_1(E)$ are products of elements of the form $\alpha\beta\overline{\alpha}\overline{\beta}=[\alpha,\beta]$. Therefore $p_*\pi_1(E)=\pi_1(X)'$. Analogously, it can be proved that the corresponding covering space of $S^1\vee S^1\vee S^1$ is $\bR\times \bZ\times \bZ\cup \bZ\times \bR\times \bZ\cup \bZ\times \bZ\times \bR$.
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