In Edixhoven, van der Geer, abelian varieties, the proof of the statement 9.18.(iii) includes the claim that for any isogeny $f : X \to Y$ of degree $k$ there is a so-called "trace map" $f_{\ast}\mathcal{O}_X \to \mathcal{O}_Y$ that induces a section of $\mathcal{O}_Y \to f_{\ast}\mathcal{O}_X$. The corollary would be that $f_{\ast}\mathcal{O}_X = \mathcal{O}_Y \oplus E$ with $E$ of rank $k - 1$. My question is:
What should we require of $X, Y$ and $f$ to make sure that $E$ actually splits in $k - 1$ line bundles belonging to $\mathrm{Pic}^0(Y)$?
It would take some time to explain the question I'm actually thinking about, so any help would be appreciated if you may.
Assume that $L$ is a line bundle on $Y$ such that $f^*L \cong O_X$. Then $$ f_*f^*L \cong f_*O_X \cong O_Y \oplus E. $$ On the other hand, by the projection formula $$ f_*f^*L \cong L \otimes f_*O_X \cong L \otimes (O_Y \oplus E) \cong L \oplus (L \otimes E). $$ So, if $L \not\cong O_Y$ then $L$ is a summand of $E$.
Thus, the condition you need is that the length $k$ subscheme $Ker (f^*) \subset Pic^0(Y)$ is a union of $k$ points defined over your base field.