Here is a (probably non-optimal) statement that may apply in your situation. In your situation with curves, the hypothesis says that you need $X$ and $Y$ to be Gorenstein.

**Claim.** *Let $X$ and $Y$ be noetherian schemes satisfying $G_1$ and $S_2$. If $f\colon X \to Y$ is a finite morphism and $\mathscr{F}$ is a coherent reflexive sheaf on $X$, then $f_*\mathscr{F}$ is a coherent reflexive sheaf on $Y$.*

*Proof.* On noetherian schemes satisfying $G_1$ and $S_2$, reflexivity is equivalent to being $S_2$ (in Hartshorne's sense) [Hartshorne 1994, Thm. 1.9]. The claim then follows since the $S_r$ property is preserved under pushforward by finite morphisms by [EGAIV$_2$, Prop. 5.7.9]. $\blacksquare$

I wanted to prove a statement for non-finite morphisms as well, and for integral schemes, you can say a bit more:

**Claim.** *Let $X$ and $Y$ be integral noetherian schemes satisfying $G_1$ and $S_2$. If $f\colon X \to Y$ is a proper dominant morphism with all fibers of the same dimension. If $\mathscr{F}$ is a coherent reflexive sheaf on $X$, then $f_*\mathscr{F}$ is a coherent reflexive sheaf on $Y$.*

*Proof.* The fact that $f_*\mathscr{F}$ is coherent and normal follows from the proof of [Hartshorne 1980, Cor. 1.7]. By [Hartshorne 1994, Rem. 1.11], to show $f_*\mathscr{F}$ is reflexive, it therefore suffices to show that it satisfies $S_1$. But being $S_1$ is equivalent to torsion-freeness for integral noetherian schemes [Hartshorne 1994, Lem. 1.5], hence the claim follows by the fact that torsion-freeness is preserved under pushforwards by dominant morphisms. $\blacksquare$