= Solution
If $f:A\to B$ has a section $g:B\to A$, functoriality gives $H_fH_g=1_{H_B}$. Thus $H_f$ is a <split epimorphism> and in particular an <epimorphism>.
For the converse, justify the needed pointwise surjectivity directly. Let $I\subseteq H_B$ be the image subpresheaf of $H_f$. Form a new presheaf by taking, at every object, two copies of $H_B(U)$ and identifying their elements in $I(U)$. Restriction maps descend because $I$ is a subpresheaf. The two canonical transformations $j_1,j_2:H_B\to H_B\amalg_IH_B$ agree after $H_f$. If any component of $I$ is proper, an element outside it distinguishes $j_1$ from $j_2$. Therefore epimorphicity forces every component of $H_f$ to be surjective.
In particular, its component at $B$ has $1_B$ in its image: there is $g:B\to A$ with $fg=1_B$. This proves the <Yoneda embedding detects split epimorphisms> criterion
$$
\boxed{H_f\text{ is epic if and only if }f\text{ is split epic}.}
$$
Ordinary epimorphicity of $f$ alone would not suffice.
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