Let $R$ be any (i.e. not necessarily commutative or unital) associative ring, and let $I$ be a (two sided) ideal of $R$. Hence $I$ is a (nonunital) ring.
How can I prove: $R$ is a Noetherian/Artinian ring iff $I$ and $R/I$ are N/A rings?
We know (Atiyah & MacDonald, p.75, prp.6.3, the proof is the same for the noncommutative nonunital case) that if $0\rightarrow M'\rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of left $R$-modules, then $M$ is N/A iff $M'$ and $M''$ are N/A.
We have an exact sequence $0\rightarrow I\rightarrow R \rightarrow R/I \rightarrow 0$ of left $R$-modules, so $R$ is N/A iff $I$ and $R/I$ are N/A as $R$-modules. Now, since every $R$-submodule of $R/I$ is a $R/I$-submodule of $R/I$, and vice versa, we know that $R/I$ is N/A as a $R$-module iff it is such as an $R/I$-module, i.e. as a ring.
How can I prove that $I$ is N/A as a $R$-module iff it is such as an $I$-module?
What you claim isn't true. Consider the ring $R=\mathbb{R}^{\mathbb{N}}$ of sequences of real numbers, with component-wise operations. This ring isn't Artinian, since the ideals $$I_n=\{(a_i)_i\in R;\forall i\leq n\colon a_i=0\}$$ form a descending chain which doesn't stabilize. It also isn't Noetherian, since the ascending chain of ideals $$J_n=\{(a_i)_i)\in R;\forall i\geq n\colon a_i=0\}$$ also doesn't stabilize.
Now consider the ideal $J_1$. It is a field, since it is isomorphic as a ring to the reals, and is as such both Noetherian and Artinian.