Is a morphism of reduced schemes over an algebraically closed field determined by its values on closed points?

If $X$ is locally of finite type and $Y$ is separated over $k$, then the answer is "yes." Indeed, the separatedness of $Y$ implies that the equalizer $Z:=\ker(f,g)\hookrightarrow X$ is a closed subscheme. The assumption that $f$ and $g$ agree on $k$-points implies that the underlying set of $Z$ contains all $k$-points of $X$. But $k$-points are nothing but closed points since $k$ is algebraically closed and $X$ is locally of finite type over $k$, so, since the closed points of $X$ are dense in $X$, the underlying set of $Z$ is $X$. Since $X$ is assumed reduced, $Z=X$ as closed subschemes, and therefore $f=g$.

If you replace "reduced" with "geometrically reduced" (in addition to the additional hypotheses I made above) then this remains true for an arbitrary $k$ (base changing along an extension of $k$ is faithful on morphisms so one immediately reduces to the case of an algebraically closed base field).

I'm not sure what happens in general, but I would guess that if $X$ is not at least locally of finite type, this won't be true.