Suppose $\mathcal E\simeq \mathsf{Sh}(\mathsf C,J)$. I know the a family $(f_i:C_i\to C)_i$ covers $C$ iff the family $(\mathbf{ay}(c_i:C_i\to C))_i$ is jointly epimorphic in $\mathcal E$, where $\bf a$ is sheafification and $\bf y$ is the Yoneda embedding.
Is it true this is in turn equivalent to the joint epimorphy of the family $(\operatorname{Im}\mathbf{ay}c_i\to \mathbf{ay}C)_i$? I know this would imply the former conditions since the arrow to the image in the epi-mono factorization is an epi, and epis are closed under composition. I'm just not sure about the converse.
Geometric intuition tells me this should hold. First of all, the arrow $\coprod_i \mathbf{ay}U_i\to \mathbf{ay}X$ is an epi iff its image is an epi, because epis have right cancellation and are closed under composition. So all that's left is to show is that images commute with coproducts.
My handwavy argument is that images in a topos are coequalizers of kernel pairs, and kernel pairs (as pullbacks) commute with coproducts (because coproducts are universal in a topos), while coequalizers also commute with coproducts. I'm just not sure how to actually prove this.
If $f_i: X_i\to Y$ is a family of morphisms each factoring as $X_i\twoheadrightarrow Z_i\rightarrowtail Y$, then $\coprod_i f_i: \coprod_i X_i\to Y$ factors as $\coprod_i X_i\to\coprod Z_i\to Y$. Hence, since coproducts of epimorphisms are epimorphisms (this follows without obstacles from their defining properties), it follows that $\text{image}(\coprod_i f_i)=\text{image}\left(\coprod_i Z_i \to Y\right)$ as wished.