The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\
\rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\
& & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0
\end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.
You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.
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.
Best Answer
Well, it turns out at the question you linked that the result isn't really about a sum of ideals. It looks like the user perhaps did not apprehend fully that every ideal of $\frac{A}{\mathfrak a}$ is of the form $\frac{\mathfrak b'}{\mathfrak a}$, and that $\mathfrak b'$ already contains $\mathfrak a$, so that $\mathfrak a+\mathfrak b'=\mathfrak b'$.
But the result you are thinking of is probably the Chinese remainder theorem, which is often found in its "most concrete" case using integers. It can actually be expressed for arbitrary commutative rings this way (as seen in I.M. Isaacs Algebra: a graduate course):
A version even holds for noncommutative rings, but if I remember correctly 26.16 breaks down for noncommutative rings.
The more general construction is to take any family of ideals $I_i$ indexed by a set $S$ and to map $r\mapsto (\ldots, r+I_i,\ldots)\in \prod_{i\in S} R/I_i$, which is always a homomorphism with kernel $K=\bigcap_{i\in S} I_i$. The problem is that it is not necessarily onto the product, and the Chinese Remainder Theorem is a special case when you can connclude it is in fact onto. In this more general construction, $R/K$ is said to be a subdirect product of the $R/I_i$'s (because it has the extra nice property that it projects onto each of the $R/I_i$'s.)
Along the lines of isomorphism theorems with intersections: you are aware of the second isomorphism theorem, right? It says that for any two ideals $A,B$ in a ring $R$, $\frac{A+B}{A}\cong \frac{B}{A\cap B}$. (In fact it works for right and left ideals, or just plain modules too, as do the other two isomorphism theorems.)