Here is a limited no answer.
All three rings $k[x,y]/(x^3,y^5)$, $k[x]/(x^3)$, and $k[y]/(y^5)$ are local, so projective = free.
The dimension of a module that is projective for both of the small rings, must be a multiple of 15, and I suspect it is clear that the action must also fit together so that the resulting module is free over the original ring.
The regular module contains no proper, nonzero submodule that is projective over both subrings (just by dimension count). Over a group ring this would mean the calculation was finished, I believe, but perhaps one has to be more careful in general. Below I give what I hope is a reduction to this case.
If the two powers are the same, then the least common multiple can be quite a bit smaller. For $k[x,y]/(xx,yy)$ one has the submodule generated by $x+y$ is free over both subrings $k[x]/(xx)$ and $k[y]/(yy)$, but of course is not free over the original (local) ring due to dimension problems.
Perhaps this calculation works can be made to work in general:
$$R = k[x,y]/(x^3,y^5) = k[x]/(x^3) \otimes k[y]/(y^5) = X \otimes Y$$
If $M$ is $X$-free, then $X\otimes M \cong M$, and if $M$ is $Y$-free, then $Y \otimes M \cong M$. Hence $$R \otimes M \cong (X \otimes Y) \otimes M \cong X \otimes ( Y \otimes M ) \cong X \otimes M \cong M$$
Something goes wrong (isomorphic over which ring?) since this doesn't use the relatively prime hypothesis, but hopefully it reduces the question to the $R$-module $R$, where it is clear in the relatively prime case.