Let $S$ be a scheme, $R,U$ be $S$-schemes and $s,t : R \to U \times_S U$ be an equivalence relation i.e. it's a monomorphisme such that for every $S$-scheme $T$, $R(T) \to U(T) \times U(T)$ is and equivalence relation in the usual sense.
We denote by $Q$ the $fppf$ sheaf associated to the presheaf $$T \mapsto U(T)/\sim_{R(T)} $$
The arrow $\pi : U \to Q$ is the coequalizer, in the category of $fppf$ sheaves, of the maps $s,t : R \to U$ seen as maps of $fppf$ sheaves. Note that as such it is an epimorphism.
Let $\pi : Q' \to Q$ be any morphism. We define $R' = R \times_Q Q'$ and $U' = U\times_Q Q'$. I want to show that $Q'$ is the coequalizer of the maps $(s',t') : R' \to U' \times_{R'} U'$. So what I want to show is that the formation of quotients of equivalence relations commutes with base change in the category of $fppf$ sheaves
I know that $(s,t) : R \to U \times_Q R$ is an isomorphism of $fppf$ sheaves (it isn't absolutely obvious but it's not very hard to see). This implies very easily that $R' \cong U' \times_{Q'} U'$. So the question becomes to show it that $U' \to Q$ is the equalizer of $p_1,p_2 : U' \times_{Q'} U' \to U'$.
My idea was to show that $U' \to Q$ is an epimorphism because I've already shown that this implies the result, but I haven't been able to show this.
So another (and maybe clearer) question would be : in the category of $fppf$ sheaves, are epimorphisms stable by base change ? Let me remind you that a morphism $\alpha : F \to G$ of $fppf$ sheaves (on the category of $S$-schemes) is an epimorphism if for every $S$-scheme $T$ and for every section $s \in G(T)$ there exists and $fppf$ covering $\{T_i \to T\}_{i \in I}$ such that $s_{|T_i}$ is in the image of $F(T_i) \to G(T_i)$.
It is certainly true in the category of sets that epimorphisms are stable by base change so I imagine the result would be true for epimorphisms of sheaves over any site.
Anyway, if it is not the case, then how can one show that $Q'$ is the co-equalizer of $(s',t')$ ?
Suppose given a parallel pair $X \rightrightarrows Y$ in $\mathbf{Sh}$. Let $Y \to \tilde{Z}$ be the coequaliser in $\mathbf{Psh}$ and let $Z$ be the sheaf associated with $\tilde{Z}$; then $Y \to Z$ is the coequaliser in $\mathbf{Sh}$. Now, consider a morphism $Z' \to Z$ and define $X' = Z' \times_Z X$, $Y' = Z' \times_Z Y$, and $\tilde{Z}' = Z' \times_Z \tilde{Z}$; note that we get a parallel pair $X' \rightrightarrows Y'$ in $\mathbf{Sh}$. Since coequalisers in $\mathbf{Set}$ are preserved by base change, $\tilde{Z}'$ is the coequaliser of $X' \rightrightarrows Y'$. Moreover, the associated sheaf functor preserves finite limits and all colimits, so $Z'$ is the sheaf associated with $\tilde{Z}'$ and is the coequaliser of $X' \rightrightarrows Y'$. Hence coequalisers are preserved by base change in $\mathbf{Sh}$.
(It may be helpful to draw some diagrams.)