Suppose $k$ is a field. I believe that $ \operatorname{Hom}_{k[x]}(k[x]/\langle x^2 \rangle , k[x]/ \langle x^3 \rangle ) = 0$, since by the universal property for quotients, homomorphisms in $ \operatorname{Hom}_{k[x]}(k[x]/\langle x^2 \rangle , k[x]/ \langle x^3 \rangle ) $ are characterised by
$$ \phi \colon k[x] \to k[x]/\langle x^3 \rangle \qquad \phi(\langle x^2\rangle)=0.$$
This means $ \phi(x^2) = 0 $ so $ x^2 \phi(1) = 0 $ and $ \phi(1) = 0 $. Therefore, $\phi(p(x)) = 0$ for any polynomial $ p(x) \in k[x] $.
But it seems we haven't really used $ k[x]/\langle x^3 \rangle $ at all, so $ \operatorname{Hom}_{k[x]}(k[x]/\langle x^2 \rangle , k[x]/ \langle x^n \rangle ) = 0$ for any $ n > 2 $?
How about $ \operatorname{Hom}_{k[x]}(k[x]/\langle x^m\rangle, k[x]/\langle x^n\rangle)$, for integers $ m, n$, is this still $0$?
This is a correction and replacement of my original response, which ignored the request for homomorphisms from $k[x]/\langle x^2\rangle$ to $k[x]/\langle x^3\rangle$ as $k[x]$-modules. I thank @Max for pointing out my error.
It helps to look at the parallel question of the homomorphisms between the $\Bbb Z$-modules $\Bbb Z/p^2\Bbb Z$ and $\Bbb Z/p^3\Bbb Z$. Of course there is a nontrivial map, $n\mapsto pn$. In the same way, we have $a+bx\mapsto ax+bx^2$, multiplication by $x$, as a nontrivial $k[x]$-homomorphism from $k[x]/\langle x^2\rangle$ to $k[x]/\langle x^3\rangle$.