Consider a curve $C$ of genus $5$ endowed with an automorphism $\sigma\colon C \rightarrow C$ of order $4$, such that the quotient of $C$ by $\sigma$ is an elliptic curve $E$ and the isotropy of any point is at most of order $2$. Let $f\colon C \rightarrow E$ be the quotient morphism. The branch locus on $E$ consists of four points $p_1,\ldots,p_4$.
Prove that there exists an invertible sheaf $\mathcal L$ on $E$ such that $\mathcal L ^4 \cong \mathcal O_E(p_1 + \ldots + p_4)^2$, $\mathcal L^2 \not \cong \mathcal O_E(p_1 + \ldots + p_4)$ and $$f_*\mathcal O_C \cong \mathcal O_E \oplus \mathcal L^{-1} \oplus \mathcal M \oplus \mathcal L^{-1}\otimes \mathcal M,$$ where $\mathcal M = \mathcal L^{-2}\otimes \mathcal O_E(p_1 + \ldots + p_4).$
Since $f$ is finite and flat, $f_*\mathcal O_C$ is locally free of rank $4$. From this answer I understand that $f_*\mathcal O_C$ should split into the sum of 4 invertible sheaves. Still, I have little clue why the invertible sheaves should look like that, although I know that this is connected to the fact that $f_*\mathcal O_C$ has the structure of an $\mathcal O_E$-algebra.
These statements were found on a paper by Kapil Paranjape, "Abelian varieties associated to certain $K3$ surfaces", Compositio Mathematica 68: 11-22 (1988), which I assume to contain some typos.