$\DeclareMathOperator{\Aut}{Aut}$ Suppose $p_1 \colon E_1 \to B_1$ and $p_2 \colon E_2 \to B_2$ are regular covering maps, with corresponding group exact sequences $1 \to \pi_1(E_i) \to \pi_1(B_i) \to \Gamma_i \to 1$ where $\Gamma_i = \Aut(p_i)$ are the deck groups.
Although $\pi_1(B_1 \vee B_2) = \pi_1(B_1) \ast \pi_1(B_2)$ (the free product) by Seifert-Van Kampen, $p_1 \vee p_2 \colon E_1 \vee E_2 \to B_1 \vee B_2$ is almost never a covering map. On the other hand, we do have a free product map $\pi_1(B_1) \ast \pi_1(B_2) \to \Gamma_1 \ast \Gamma_2$ which is surjective.
Is there a topological way to construct the regular cover of $B_1 \ast B_2$ with deck group $\Gamma_1 \ast \Gamma_2$ using just the maps $p_1$ and $p_2$ (that is, without identifying the kernel and going to the universal cover)? In particular, I would like a description in the case where $B_1 = B_2 = S^1$.
Note that this is different than asking for a covering map which realizes the inclusion $\pi_1(E_1) \ast \pi_1(E_2) \to \pi_1(B_1) \ast \pi_1(B_2)$, although that might me more suitable to be called the free product of covering maps. On the other hand, this cover would not be regular in general, while the one I'm asking for is.