You actually already proved the difficult part.
If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.
Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.
Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.
The ideals of $\mathbb{Z}_n$ are, first of all, additive subgroups of $\mathbb{Z}_n$. These we know to all have the form $\langle d\rangle$, where $d$ divides $n$. But, as we know, the set $\langle d\rangle$ is the ideal generated by $d$. So we have just proven that
the ideals in $\mathbb{Z}_n$ are precisely the sets of the form $\langle d\rangle$ where $d$ divides $n$.
Since we are interested in maximal ideals, and this concept is defined in terms of containment
of ideals in one another, we now need to determine when we can have $\langle d_1\rangle\subset \langle d_2\rangle$. This is the case if and only if $d_1 \in <d_2>$.
Here is the main result that you are seeking for:An ideal $I$ in $\mathbb{Z}_n$ is maximal if and only if
$I = \langle p \rangle$ where p is a prime dividing n.
Best Answer
If $f(M) \subseteq I \subseteq S$ is an ideal, then $M \subseteq f^{-1}(I) \subseteq R$. Since $M$ is maximal, we get $M=f^{-1}(I)$ or $f^{-1}(I)=R$, i.e. $f(M)=I$ or $I=S$. $\mathrm{QED}$