Let $X$ be a topological space. If $A \leq B$ are Abelian group-valued sheaves on $X$, then the presheaf quotient $U \mapsto B(U)/A(U)$ is a separated presheaf.
Does this still hold if we require $A$ and $B$ only to be separated presheaves?
I believe the answer is "no". In this case, could you please provide a counterexample?
The obvious proof I can think of for the original statement will still work only assuming $A$ is a sheaf, and $B$ is a separated presheaf. More generally, it will work if $A$ is a closed subpresheaf of $B$ (where this means that whenever $x \in B(U)$ and we have a open cover $\{ V_i \mid i \in I \}$ of $U$ such that $x |_{V_i} \in A(V_i)$ for each $i$, then $x \in A(U)$). (Note that we have the general result that if $A$ is a subpresheaf of $B$, $A$ is a sheaf, and $B$ is separated, then $A$ is closed in $B$; the straightforward proof of this fact might very well have been implicit in the proof of the statement you started out with.)
Thus, for a counterexample where $A$ and $B$ are separated, we will need a situation where $A$ is not closed in $B$. For one such counterexample, let us suppose $X = \mathbb{R}$, and we take $B$ to be the constant presheaf $\mathbb{Z}$, while $A$ is the subpresheaf $\mathbb{Z}_{(-\infty, 1)} + \mathbb{Z}_{(-1, \infty)}$. To be more explicit, this is the subpresheaf where $n \in \mathbb{Z} = B(U)$ is in $A(U)$ if and only if either $n = 0$, $U \subseteq (-\infty, 1)$, or $U \subseteq (-1, \infty)$. Now for example consider the global sections $1, 0 \in B(\mathbb{R})$. Then $\bar 1 |_{(-\infty, 1)} = \bar 0 |_{(-\infty, 1)} \in (B /_{psh} A)(-\infty, 1)$ and $\bar 1 |_{(-1, \infty)} = \bar 0 |_{(-1, \infty)}$; however, $\bar 1 \ne \bar 0 \in (B /_{psh} A)(\mathbb{R})$.
(You might even see a way to generalize the above argument to a converse of the above observations: if $B /_{psh} A$ is separated, then necessarily $A$ must be closed in $B$.)