Let $S$ be the set of all subgroups $A$ of $G$ such that $N\subseteq A$. Let $T$ be the set of all subgroups of $G/N$. The claim is that there is a bijection between the sets $S$ and $T$. The use of the word bijection is the usual one: there exists a function $\psi:S\to T$ which is both injective and surjective.
Side remark: The theorem is much stronger. Each of the sets $S$ and $T$ above are in fact lattices and the bijection is a lattice isomorphism. Moreover, the isomorphism preserves and reflects normality.
Applications:
1) A normal subgroup $N\subseteq G$ is maximal (i.e., no subgroup $H$ exists with $N\subseteq H \subseteq G$) iff the index of $N$ in $G$ is a prime number.
2) Let $N$ be normal in $G$. Then $G/N$ is simple iff there exists no normal subgroups $K$ of $G$ with $N\subset K \subset G$.
A similar result holds for rings, giving the classical and very important application
3) If $I$ is an ideal in a commutative ring $R$, then $R/I$ is a field iff $I$ is maximal in $R$. This gives a very efficient way to building fields.
Answer to you later edit: Yes, normality is both preserved and reflected by the bijection. That means that if $\psi(A)=H$, then $A$ is normal in $G$ iff $H$ is normal in $G/N$. Most other properties of groups though will not be preserved or reflected in this way.
This answer includes an application of the 4th isomorphism theorem for rings, and one (well, kinda two) of the 2nd isomorphism theorem for groups. I also link to a really cool application of the 4th isomorphism theorem in groups (but is kinda the same as the application to rings).
Rings, 4th isomorphism theorem
An neat application of the 4th isomorphism theorem for rings is the following:
Theorem: Maximal ideals are prime.
Proof: Let $R$ be a ring and let $I$ be a maximal ideal of $R$. Then consider the quotient $R/I$, and apply the correspondence theorem.
Indeed, if $R$ is commutative with identity then $R/I$ is a field. See here for a proof.
Groups, 4th isomorphism theorem
There is a rather cool trick in the theory of infinite groups, which was used by Higman to construct an infinite simple group. The idea is to appeal to Zorn's lemma and obtain a maximal normal subgroup, and then quotient this out to get a simple group. See here for a neat application of it. This is very similar to the above application to rings. (You quotient out a maximal subgroup/ideal to get a simple group/prime ring.)
Groups, 2nd isomorphism theorem
A nice application of the 2nd isomorphism theorem for groups is the following. It deals with soluble groups, which are an important class of groups (and are often taught in a second course on groups). A group $G$ is soluble (or solvable) if it possesses an abelian series, that is, a series $$1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$$ where each factor $G_{i+1}/G_i$ is abelian.
Theorem: Subgroups and homomorphic images of soluble groups are soluble.
Proof: Suppose $G$ is soluble with abelian series $1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$.
Subgroups: If $H$ is a subgroup of $G$ then we can apply the 2nd isomorphism theorem to get the following.
$$\frac{H\cap G_{i+1}}{H\cap G_i}\cong\frac{(H\cap G_{i+1})G_i}{G_i}\leq\frac{G_{i+1}}{G_i}$$
Hence, the groups $\frac{H\cap G_{i+1}}{H\cap G_i}$ are abelian so the set $\{H\cap G_i; i=0, 1, \ldots, n\}$ forms an abelian series for the group $H$, as required.
Homomorphic images: If $N$ is a normal subgroup of $G$ then the can apply the 2nd isomorphism theorem to get the following.
$$\frac{G_{i+1}N}{G_iN}\cong\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$$
The subgroup $G_i$ is a subgroup of $G_{i+1}$ and of $G_iN$, and so $G_i\lhd G_{i+1}\cap (G_iN)$. Therefore, $\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$ is a homomorphic image of $\frac{G_{i+1}}{G_i}$, and hence is abelian. By the 3rd isomorphism theorem we have the following.
$$\frac{G_{i+1}N/N}{G_iN/N}\cong \frac{G_{i+1}N}{G_iN}$$
Hence, the set $\{G_iN/N; i=0, 1, \ldots, n\}$ forms an abelian series for $G/N$, as required.
Best Answer
Let's be clear about what the bijection is between. There is a 1-1 correspondence $$ \{\mbox{Ideals $J$ s.t. $I\subseteq J\subseteq A$}\}\leftrightarrow\{\mbox{Ideals in $A/I$}\} $$ This bijection is induced by the canonical projection $\pi:A\to A/I$.
Note that if $J$ is an ideal containing $I$, then $\pi(J)$ is an ideal in $A/I$. Also, if $K\lhd A/I$ is an ideal, then $\pi^{-1}(K)=\{x\in A\mid \pi(x)\in K\}$ is an ideal of $A$ containing $I$.
To see that this gives you a bijection, it is enough to prove two things:
If $K\lhd A/I$ is an ideal, then $\pi(\pi^{-1}(K))=K$. (This is by definition).
If $J$ is an ideal containing $I$, then $\pi^{-1}(\pi(J))=J$. (It is clear that $J\subseteq\pi^{-1}(\pi(J))$. To get the reverse inclusion, suppose $a\in\pi^{-1}(\pi(J))$. Then $a+I=j+I$ for some $j\in J$. If follows that $a-j\in I\subset J$, forcing $a\in J$.)
Now, the next statement is that this bijection preserves inclusions. That is, if $I\subset J_1\subset J_2\subset A$ are two ideals, then $\pi(J_1)\subset\pi(J_2)$.
This is called the Lattice Isomorphism Theorem because the ideal lattice for $A/I$ is the same as the portion of the ideal lattice for $A$ with minimal element $I$.