What makes this theorem fundamental, and why is it an "isomorphism theorem"? Theorem: Let $A$ be a commutative ring with ideal $I$. Then, define the quotient homomorphism $\phi: A \to A/ I$ given by $a \mapsto a+I$. Then $\phi^{-1}: \{ \textrm{ideals of $A/I$}\} \to \{\textrm{ideals of $A$ containing $I$}\}$ is an inclusion-preserving bijection.
I've seen some proofs of this theorem, none of which make a lot of sense to me, which can be found here or here. My attempted proof is more or less as follows: define another quotient surjection $\rho:A/I \to (A/I)/J$, where $J$ is an ideal of $A/I$. Then
$$\ker(\rho \circ \phi)=\phi^{-1}(\ker \rho)=\phi^{-1}(J)$$
implying that $J$ is the kernel of a surjective homomorphism, and hence an ideal of $A$, that clearly contains $I$, since $\ker\phi \subseteq\ker  \rho \circ \phi$.
After proving the theorem, I tried to work out an example to try to get a sense for what it was really telling me:
Consider $\mathbb Z$, and let $(x)$ denote the ideal generated by $x \in \mathbb Z$.Let $I=(6)$. Then the ideals of $\mathbb Z$ contain $I$ are: $(1),(2),(3), (6)$. Ideals of $\mathbb Z/6\mathbb Z$ are $(1)=(5)=\mathbb Z/6\mathbb Z$, $(2)=(4)=\{2,4,0\}$, $(3)=\{0,3\}$, and finally $(0)=\{0\}$. Checking the preimages of these ideals, I got exactly $(1),(2),(3), (6)$, so I felt good about this.
I've seen a question here giving some applications of the theorem.
Yet, the applications in some sense seem like technical results, I don't really understand what this theorem is saying. I can't decipher why it may be important, and I can't think of a proof that isn't just symbol pushing. 


*

*Is there any way to understand why this theorem is fundamental, or what it is really saying?

*What is the isomorphism here? There is a $1-1$ correspondence, but I don't see what the isomorphism, or why it's of interest?
edit why does this theorem imply that if $I,J$ are ideals so that $I \subseteq J$, then $(A/I)/(J/I) \cong A/J$?
 A: As a matter of notation in what follows I use $J \leq A$ to mean $J$ is an ideal of $A$.
The theorem basically establish the existence of a two bijective maps, one inverse to each other, between the sets
$$\{J \subseteq A \colon J \leq A, J \supseteq I\}$$
and
$$\{J'\subseteq A/I \colon J' \leq A/I\}\ .$$
The importance of these bijection is given by the fact that they are monotone and preserve many property of the ideals: for instance the bijections restricts to bijections between the set of prime and maximal ideals.
The importance of the theorem comes from the fact that it allows to:


*

*reduce the study of ideals of a quotient ring $R/I$ to the study of the ideals of $R$ containing $I$

*reduce the study of ideal of $R$ containing an ideal $I$ to the study of the ideals of the ring $R/I$.


This is important because sometimes it happens that we have a good knowledge or understanding of just one of the rings $R$ or $R/I$ but we are interested in studying properties of the ideal of the other ring. The theorem allows to find out the wished properties by looking to the ring we know the best.
Now what are these two bijection? Well they are simply the mappings
$$\phi \colon \{J \subseteq A \colon J \leq A, J \supseteq I\} \to \{J'\subseteq A/I \colon J' \leq A/I\}$$
$$\phi(J)=\{j+I \colon j \in J\}$$
and 
$$\phi^{-1} \colon \{J'\subseteq A/I \colon J' \leq A/I\} \to \{J \subseteq A \colon J \leq A, J \supseteq I\}$$
$$\phi^{-1}(J')=\{j \in A \colon j+I \in J'\}\ .$$
Of course one have to prove that both $\phi(J)$ and $\phi^{-1}(J')$ are indeed ideals (above they are defined just as sets) and that indeed the equalities of functions $\phi^{-1}\circ\phi=\text{id}$ and $\phi\circ\phi^{-1}=\text{id}$ both hold, but that is just a matter of calculations and applying the definitions.
Hope this helps.
Edit: (about the third isomorphism theorem).
Let $A$ be a ring and $I \leq J \leq A$ be ideals. As a matter of notation we can denote by $J/I$ the ideal $\phi(J)$. 
The third isomorphism theorem tells that there is an isomorphism 
$$A/J \cong (A/I)/(J/I)\ .$$
How does this theorem relate to the ideal-correspondence described above?
For start we can consider the homomorphisms $\pi \colon A \to A/I$ and $\pi' \colon A/I \to (A/I)/(J/I)$. These homomorphisms give us 
$$\varphi = \pi' \circ \pi \colon A \to (A/I)/(J/I)$$
and the first homomorphism theorem we have an isomorphism 
$$\bar \varphi \colon A/\ker \varphi \to (A/I)/(J/I)\ .$$
We can easily notice that 
$$\ker \varphi = (\pi' \circ \pi)^{-1}((0))=\pi^{-1}(\pi'^{-1}((0)))=\pi^{-1}(J/I)$$
but by definition $J/I=\phi(J)$ and so we have also that 
$$\ker \varphi=\pi^{-1}(J/I)=\phi^{-1}(J/I)=J$$
and so 
$$\bar \varphi \colon A/J \to (A/I)/(J/I)$$
q.e.d.
