Theorem : In a regular category $C$, if
$\require{AMScd}$ \begin{CD} A@>a>>X\\ @VfVV@VVgV\\ B@>>b>Y \end{CD}
is a pullback diagram and $b=c_1c_2$ with $c_1$ and $c_2$ regular epis, say $c_1:C\to B$ is the coequalizer of $P\overset{p_0 }{\underset{p_1}\rightrightarrows}C$ and $c_2:C'\to C$ is the coequalizer of $Q\overset{q_0 }{\underset{q_1}\rightrightarrows}C'$ then $a=a_1a_2$ for some regular epis $a_1,a_2$.
Proof : We first put all the information in the commutative diagram.
Doubt : If there are morphims $a_2: Q \to P$ and $a_1: P \to A$ such that the diagram commutes, we are done because regular epi's are stable under pullback. I struggle to prove those arrows exist. As a regular category contains all finite limits, the limit of $C \xrightarrow{c_1} B \xleftarrow{f} A$ must exist, which is a pullback square, but we cannot choose $P$ to be the final vertex of the square. Call that final vertex $Z$, I think we have to use the universal property of a pullback square, but I don't know how.
Consider the following commutative diagram, where both squares are pullback. $\require{AMScd}$ \begin{CD} A'@>a_2'>>C'@>a_1'>>X\\ @Vf'VV@VVV@VVgV\\ B@>>c_2>C@>>c_1>Y \end{CD} By pullback pasting lemma, the outer square is a pullback as well. Consequently, there exists one and only one isomorphim $h:A\to A'$ such that $f=f'h$ and $a=a_1'a_2'h$. As you pointed out, $a_i'$ are regular epimorphism. Since $a_2'h$ is a regular epimorphism as well, this gives the required decomposition.