Let $I$ be a left ideal of ring $R$. My question is can we lift both Noetherian and Artinian property of $R/I$ as an $R$-module to $R$ as an $R$-module?
Precisely, can we say that $R$ is a left Noetherian (Artinian) $R$-module if and only if $R/I$ is left Noetherian (Artinian) $R$-module?
One implication If $R$ as left $R-$module is (Noetherian) Artinian then $R/I$ is too. Suppose it is not, suppose we have infinite (ascending) descending chain of modules in $R/I$ then we have the same (ascending) descending chain of submodules in $R$.
(Recall that lattice (poset) of left ideals containing $I$ is the same as a lattice of submodules in cyclic $R-$module $R/I$ by correspondence theorem.)
This is true even more generally for any lattice. If there are no infinite (ascending) descending chains in lattice $\mathcal{L}$ then there can't be such chains in the sublattices defined as all elements that are above some element $I$.
Second implication As noted in comments there are some trivial cases when it fails, like $R/R$ is always Artinian for any ring $R$.
However, (and this is special for modules, i.e. for Artinian/Noetherian property not true in general lattices), if $M$ is $R-$module such that $N$ and $M/N$ are (Noetherian) Artinian submodules, for some $R-$submodule $N$ then $M$ itself is (Noetherian) Artinian. (and as a consequence for any submodules $K$ it holds that $K$, $M/K$ are Artin.)