We know the following fact: Noetherian submodule $N$ of $A$-module $M$ is also Noetherian.
Indeed, consider an exact sequence $0\to N \xrightarrow{f} M \xrightarrow{g} M/N\to 0$. Since $M$ is Noetherian then $N$ is also Noetherian.
But subring of Noetherian ring may not be Noetherian. For example, if $\mathbb{k}$ is field then $\mathbb{k}[x_1,x_2,\dots]$ is not Noetherian but it is a subring of field of fractions which is Noetherian.
But I was wondering what if we make reasoning by the following way: Let $R$ be Noetherian ring and $S$ is a subring of $R$. Then we can consider the ring $R$ as $R$-module and subring $S$ as $S$-submodule (NOT $R$-submodule)!
Am I right that this is the reason why we cannot use the above fact?
Perhaps tje point you’re trying to make is this?
A ring $R$ is a noetherian ring iff it’s a noetherian $R$-module.
A subring $S$ may not be noetherian.
But, If $S$ happens to also be an $R$-submodule (note: most subrings aren’t submodules) then it is noetherian as an $R$-module.
This is not a contradiction, because noetherianness as an $R$ module is not the same as noetherianness as an $S$-module.
If so, yes that’s right.