Let $\pi:X\to Y$ be a finite morphism of smooth projective curves over an algebraically closed field (of characteristic zero if necessary) which are both of genus $>1$. We have two "natural" maps on the level of Jacobians: $$\pi_\ast : J_X \to J_Y,\quad \pi^\ast : J_Y\to J_X$$ which satisfy $$\pi_\ast \pi^\ast = [\deg \pi],$$ where $[\deg \pi]$ is the multiplication by $\deg \pi$ map on $J_Y$. (Does this make sense?)
Let $K$ be the kernel of $\pi_\ast$. This is an abelian variety.
I'm a bit confused. Clearly, $J_X$ is isogenous to $J_Y \oplus K$.
But, is the degree of the isogeny from $J_X$ to $J_Y \oplus K$ given by $\deg \pi$ or $(\deg \pi)^2$?
First of all, $\ker\pi_*$ isn't always an abelian subvariety; you have to take the connected component of $\pi_*$ that contains 0. You may wish to assume that $\pi^*$ is injective (which is the case if $\pi$ doesn't factor through a cyclic étale covering of degree $\geq 2$), since then $\ker\pi_*$ is actually connected and everything works out a little bit better.
Let $N_K$ be the norm endomorphism of $K$, and let $\phi:J_X\to \pi^*J_Y\times K$, where $x\mapsto(\pi_*(x),N_K(x))$ (is this the morphism you're asking about?)
When $\ker\pi_*$ is connected, $\ker\phi$ is just $K\cap\ker N_K=K[d]$ ($d$-torsion points of $K$), where $d$ is the exponent of $K$ (the exponent of $K$ is the exponent of the finite group $K\cap\pi^*J_Y$), which in this case is just $\deg\pi$. Therefore $\phi$ has degree $(\deg\pi)^{2\dim K}=(\deg\pi)^{2(g_X-g_Y)}$.