I have some specific questions regarding this not answered here: Are Dummit and Foote making a mistake in proving Cohen's theorem? (for example).
Note, I don´t claim the exposition below is a full proof. I am only trying to show (and get feedback) on the things I am struggling with.
a) If we want to show that $R/I$ is noetherian, can one reasons as follows: We know that all ideals in $R/I$ are on the form $J/I$ for $I \subseteq J \subseteq R$. Now, by maximility of $I$ among non-finitely generated ideals, all such ideals $J$ are finitely generated.
Now, I´ll use the following lemma 1: If $J$ is finitely generated then $J/I$ is finitely generated.
Proof: Assume that $J = \langle j_1,\ldots,j_n \rangle$ as an $R$-module. Let $\varphi:I \twoheadrightarrow J/I$ be a surjective $R$-module homomorphism. One can observe that if $J \subset R$ is an ideal $\iff$ $J$ is an $R$-submodule of $R$ (where $R$ has a canonical $R$-module structure, that is, ordinary multiplication and addition, I believe). Then if $I \subset J$ is a submodule of $J$, then $J/I$ has a natural $R$-module structure given by $r \cdot (j+I) = rj+I$. Now, we want to show that $\varphi$ is an $R$-module homomorphism. $\varphi(rj) = rj+I = r \cdot (j+I) = r \varphi(j)$ follows from the $R$-action on $J/I$. Furthermore, $\forall k + I \in J/I, \exists p \in J$ so that $\varphi(p) = k+I$. Then $\exists r_1,\ldots,r_n \in R$ so that $r_1j_1+ \ldots+r_nj_n = p$. Then $$r_1\varphi(j_1)+ \ldots + r_n\varphi(j_n) = k+I.$$ Hence $J/I$ is finitely generated by $\langle \varphi(j_1),\ldots,\varphi(j_n) \rangle$.
By the contrapositive, if $J/I$ is not finitely generated $\implies$ J is not finitely generated (contradiction!). Hence all ideals $J/I \subset R/I$ are finitely generated $\implies R/I$ is noetherian.
$c)$ First, one want´s to show that we can give $J_1/J_1J_2$ an $R/I$-module structure. As in $a)$, we have that $J_1J_2 \subset R$ is an $R$-submodule, hence it inherits an $R$-action from the canonical R-module structure on $R$-itself. Now, $J_1J_2 \subset J_1$ and then we can give $J_1/J_1J_2$ an $R$-module action.
We have the following diagram 
Here, we get $\phi \circ q = \varphi$ so that $\varphi(r) = r+J_1J_2 = \phi(r+I)$. Then we can give the action $R/I \times J_1J_2 \to J_1J_2$ by $$r+I \cdot (j+J_1J_2) = \phi(r+I) \cdot j+J_1J_2 = (r+J_1J_2) \cdot (j+J_1J_2) = rj+J_1J_2.$$
Now, since $I \subset J_1$ we have that $I/J_1J_2$ is an $R/I$-submodule of $J_1/J_2J_2$ if I am not mistaken, and $I/J_1J_2$ is well-defined since $I,J_1J_2$ are ideals in $R$ where $J_1J_2 \subset I$ and $R$ is commutative, so as additive abelian groups, $J_1J_2$ is a normal subgroup of $I$.
We will use the following lemma 2: If $R$ is a noetherian ring, then $M$ is a noetherian $R$-module if and only if $M$ is finitely generated as an $R$-module.
Switch $R = R/I$ in the lemma, and since $J_1/J_1J_2$ is an $R$-module, where $J_1$ is finitely generated, we must have that $J_1/J_1J_2$ is finitely generated by lemma $1$. Since $I/J_1J_2$ is an $R/I$-submodule of $J_1/J_1J_2$, we find that $I/J_1J_2$ must be finitely generated.
That is, $\forall (i +J_1J_2) \in I/J_1J_2$ we have $(k_1,\ldots,k_n) \subset I$ so that $r_1k_1+ \ldots+r_nk_n + J_1J_2 = i+J_1J_2$, where $r_i \in R \setminus{I}$. But this means that $$\langle j_1,\ldots,j_n,k_1,\ldots,k_n \rangle$$ generates $I$ (with coefficients from $R$), where $\langle j_1,\ldots,j_n \rangle.$ Hence $I$ is finitely generated (contradiction!).
Should one here draw the conclusion that the collection of ideals that are not finitely generated must be empty?