Suppose I am given a space $S$, and let some new space be the wedge $X=S\vee S$. Suppose I know $\rho: \tilde{X} \rightarrow X$ is a universal abelian cover, for the space X which preserves the base point, that is $\rho(\tilde{x}_{0})=x_{0}$. Then it follows that $\rho \ast : \pi_{1}(\tilde{X},\tilde{x_{0}}) \rightarrow \pi_{1}(X,x)$ is a monomorphism.
Furthermore I know that there is a bijective correspondence between the cosets of the quotient $\pi_{1}(X)/ \rho \ast \pi_{1}( \tilde{X})$ and the preimage of the base point with respect to $\rho$, that is the set $\rho^{-1}(x_{0})$.
How can I compute $\pi_{1}(X)/ \rho \ast \pi_{1}( \tilde{X})$ using such universal abelian cover?
I am really struggling with understanding this and I think it would be helpful if someone could give me a guided example.
So as I mentioned in the comments, I think the punchline comes from the following fact:
(*) : such that usual covering theory applies, this is clearly the case of $\mathbb RP^2\vee \mathbb RP^2$
This theorem is the main story of covering space theory (although there are other, better ways to phrase it), as it essentially says that studying subgroups of $\pi_1(X,x)$ is the same as studying covering spaces of $X$.
Proving it essentially relies on the lifting theorem for covering spaces.
Once you have that, the computation you want to do is pretty straightforward : if you have an abelian covering $\tilde X\to X$ (say $X$ is based at $x$), which means that it is normal (i.e. the subgroup associated to $\tilde X$ does not depend on the choice of a basepoint $b\in p^{-1}(x)$) and that its group of automorphisms (here, $\pi_1(X)/p_*\pi_1(\tilde X)$) is abelian; then $\pi_1(X)/p_*\pi_1(\tilde X)$ is abelian, so $p_*\pi_1(\tilde X)$ contains $[\pi_1(X),\pi_1(X)]$, the commutator subgroup of $X$.
This is the smallest normal subgroup $H$ of $\pi_1(X)$ such that $\pi_1(X)/H$ is abelian.
In particular, since we have an order preserving bijection between based covering maps and subgroups, if we take a covering map $\rho : Y\to X$ which corresponds to the commutator subgroup itself (which is normal), then $\pi_1(Y)\subset \pi_1 (\tilde X)$, and so there is a map of based connected covering maps $(Y,b)\to (\tilde X, c)$ (for any choices of $b,c$, since we chose normal coverings)
This means that $(Y,b)$ is the universal abelian cover of $(X,x)$.
In particular, $\rho_*\pi_1(Y) = [\pi_1(X),\pi_1(X)]$ and so $\pi_1(X)/\rho_*\pi_1(Y) = \pi_1(X)/[\pi_1(X),\pi_1(X)] = \pi_1(X)^{ab}$, the abelianization of $\pi_1(X)$ (this is the largest abelian quotient of $\pi_1(X)$)
Now there are various ways of computing this.
if you know van Kampen's theorem, and if $S$ is nice enough (here, it's $\mathbb RP^2$, so that's the case), you can compute $\pi_1(S\vee S) = \pi_1(S)*\pi_1(S)$ (the free product of $\pi_1(S)$ with itself); and then you can check by hand that $(G*H)^{ab} = G^{ab}\times H^{ab}$ (using the definition as "largest abelian quotient" for instance). So here, you get $\pi_1(S)^{ab}\times \pi_1(S)^{ab}$, or in the specific case $S= \mathbb RP^2$, so $\pi_1(\mathbb RP^2) =\mathbb Z/2$, you get $\mathbb Z/2\times \mathbb Z/2$.
If you know homology, then you'll recognize $\pi_1(X)^{ab}$ from another theorem : Hurewicz's theorem says that for connected $X$, $\pi_1(X)^{ab}= H_1(X)$, so here $\pi_1(S\vee S)^{ab} = H_1(S\vee S) = H_1(S)\oplus H_1(S) = \pi_1(S)^{ab}\times \pi_1(S)^{ab}$, and so you may conclude as above.