I don't remember what an exact category is in this context, but in any case this seems to be true in any pointed category. Name the arrows $i:A\to B$, $p:B\to C$, $j:B'\to B$, and $q:B\to B''$, and suppose that $pj$ is epi. If $fqi=0$ then by the universal property of $p:B\to C$ we have $fq=gp$ for a unique $g$. Now $gpj=fqj=f0=0$ and $pj$ is epi so $g=0$. Thus $fq=gp=0$, but $q$ is epi so $f=0$. This proves that $qi$ is epi.

I don't remember what an exact category is in this context, but in any case this seems to be true in any pointed category. Name the arrows $i:A\to B$, $p:B\to C$, $j:B'\to B$, and $q:B\to B''$, and suppose that $pj$ is epi. If $fqi=0$ then by the universal property of $p:B\to C$ we have $fq=gp$ for a unique $g$. Now $gpj=fqj=f0=0$ and $pj$ is epi so $g=0$. Thus $fq=gp=0$, but $q$ is epi so $f=0$. This proves that $qi$ is epi.

Edit: the comments below point out that this only shows that if $fqi=0$ then $f=0$. So this argument, as is, would need the category to be Ab-enriched. If instead it is exact, in the sense specified in the comments, then to finish I should prove:

Lemma: Let $h$ be an arrow with the property that if $fh=0$ then $f=0$. Then $h$ is epi.

Proof: The assumptions tell us that the cokernel of $h$ is $0$. Factorize $h$ as $me$, with $e$ epi and $m$ mono. Since $e$ is epi, we have $cok(m)=cok(me)=0$. But $m$ is a mono, so is the kernel of its cokernel, in this case 0, so $m$ must be invertible. This now proves that $h$ is epi.