An right $R$ module is, as you said, an abelian group $N$ and the map from $N\times R\to N$ satisfying special properties like distributivity etc. In other words, $nr$ is another element of $N$ for every $n\in N$, and $r\in R$.
If $N'$ is an additive subgroup of $N$, we say that it's a submodule if $n'r\in N'$ for every $n'\in N'$ and $r\in R$. In other words, it looks like $N'R\subseteq N'$ (which didn't necessarily have to happen!).
Now $R$ can be considered as a right module over itself, since the ring multiplication $R\times R\to R$ satisfies all the axioms that are necessary to make $R$ into a right module. Since we have already defined what a submodule is, we can ask what the submodules of the right module $R$ are. They an additive subgroup $I$ of $R$ is a submodule if $IR\subseteq I$.
You'll notice that looks like the definition of a right ideal: and that's exactly what it is! The term "right ideal" is just another term for a submodule of the right module $R$. Similarly, $R$ can be considered as a left module over itself, and you can ask about its left submodules (=left ideals). Nobody says "ideals in a module" because ideals are reserved as a term for submodules of $R$.
If you already know that for any submodule $N'$ of $N$ you can form the quotient module $N/N'$, then it will be no surprise that if $M$ is a right ideal (=right submodule!) of $R$, then you can form the quotient module $R/M$.
The proof of the title fact is very easy if you know the isomorphism theorems for modules. If $S$ is a right $R$ module (nonzero, of course), pick a nonzero $s\in S$ and make a map from $R\to S$ by $r\mapsto sr$. Call this map $\phi$. This is clearly a nonzero map to $S$. Since the image is a submodule of $S$ and $S$ is simple, it must be all of $S$. By the isomorphism theorem, $R/\ker(\phi)\cong Im(\phi)=S$. By the correspondence of submodules, the absence of submodules of $S$ corresponds to an absence of submodules between $R$ and $\ker(\phi)$, and so $\ker(\phi)$ is a maximal right ideal of $R$.
Best Answer
You are correct. It isn't in general an ideal. An easy way to see this is that if your rings are unital, then the image of a homomorphism must contain 1, so it could only be an ideal if the homomorphism were surjective.
Alternatively pick any subring that is not an ideal and let your homomorphism be the inclusion.
Another alternative way to see this is what you mentioned in your post.