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 $R$.
Assume not. Then because every ideal is contained in a maximal ideal. If $\ker f$ is not maximal, then $(\ker f)\subset M_0\subset R$ with $M_0$ maximal, and $M_0/(\ker f)$ is a non-trivial ideal of $R/(\ker f)$, contradicting simplicity of $M$.
Best Answer
Since the map $\phi_m$ is surjective we have $M=Rm$ for all $m\in M-\{0\}$. Suppose $\mathrm{Ann}(m)$ is not left maximal for some $m\in M-\{0\}$. Then there is $\mathrm{Ann}(m)\subsetneq I\subsetneq R$ a left ideal. Let $a\in I-\mathrm{Ann}(m)$. Then $am\ne 0$, and $M=R(am)$. Since $M=Rm$ we get $m\in R(am)$, so there is $r\in R$ such that $m=ram$, that is, $(ra-1)m=0$ which implies $ra-1\in\mathrm{Ann}(m)$, hence $ra-1\in I$. But $ra\in I$, so $1\in I$, a contradiction.