What does the correspondence of ideals containing $I$ and ideals of $R/I$ tell us about computing quotients? $\newcommand{\Z}{\mathbb{Z}}$
Here is the situating example.
Example: Let $R = \Z[x]$ and consider the ideal $I = (3,x^2-13) = \{3f+(x^2-13)g \mid f,g \in R\}$. The goal is to compute the quotient $R/I$ and to do it we leverage the Third Isomorphism Theorem for rings.

3rd Iso. Theorem: Let $I \subseteq J$ be two-sided ideals in a ring $R$. Then $J/I$ is a two-sided ideal in $R/I$ and $(R/I)/(J/I) \cong R/J$. Moreover $J \mapsto J/I$ defines a bijection between the set of ideals of $R$ containing $I$ and the set of two sided ideals of $R/I$.

Let $J = 3R \subseteq I$, then by the 3rd Iso. Theorem we have that $R/I \cong (R/J)/(I/J)$ and so we want to compute the two quotients. To compute $R/J$ consider the map $\varphi: \Z[x] \rightarrow (\Z/3\Z)[x]$ which acts by reducing coefficients modulo $3$. This is a surjective ring homomorphism and has kernel equal to $J$ and so by the 1st Iso. Theorem we conclude that
$$
\Z[x]/(3\Z[x]) = R/J \cong (Z/3\Z)[x].
$$
To compute the second quotient is where I have issues which I believe stem from not understanding the correspondence portion of the 3rd Iso. Theorem. The notes say that computing $I/J$ corresponds to computing $\varphi(I)$ where $\varphi$ is the reduction homomorphism. Why is this the case?
When the theorem says that $J \mapsto J/I$ defines a bijection what is this map? Are we saying that this is the natural projection map?
How can I relate this to the correspondence isomorphism theorem for groups (which I assume it is related?) In that case if $G$ is a group and $N \unlhd G$ then every subgroup of $G/N$ is of the form $H/N$ for some $H \leq G$ containing $N$, but this stems from utilizing the natural projection map $\pi:G \rightarrow G/N$ by $g \mapsto gN$. I dont believe that $\varphi$ in this case is the natural projection map right?
Thanks in advance for the help, let me know if I can clarify more, I concede my thoughts are pretty jumbled.
 A: 
When the theorem says that $J \mapsto J/I$ defines a bijection what is this map?
Are we saying that this is the natural projection map?

I’m gonna switch the roles of $I$ and $J$, to make it compatible with the rest of your question.
(The isomorphism theorem uses $I ⊆ J$, but in your specific example you have $I ⊇ J$ instead.)
Let $R$ be any ring and let $J$ be an ideal of $R$.
Let $π \colon R \to R/J$ be the canonical quotient homomorphism.
We have the two sets
$$
  \mathcal{I} = \{I ⊆ R \mid \text{$I$ is an ideal of $R$ with $I ⊇ J$}\} \,,
  \quad
  \mathcal{K} = \{K ⊆ R/J \mid \text{$K$ is an ideal of $R/J$}\} \,.
$$
If $I$ is an ideal of $R$ that contains $J$, then $I / J$ (the quotient taken on the level of abelian groups) is a subset of $R/J$.
It is, in fact, an ideal of $R/J$.
We have therefore the well-defined map
$$
  \mathcal{I} \longrightarrow \mathcal{K} \,,
  \quad
  I \longmapsto I/J \,.
$$
This map is a bijection.
This map is not the canonical quotient map $π \colon R \to R/J$, but it is (in some sense) induced by it.
Note that

*

*by the third isomorphism theorem, both the original ideal $I$ of $R$ and the induced ideal $I/J$ of $R/J$ result in isomorphic quotient rings:
$$
  (R / J) / (I / J) ≅ R / I \,;
$$


*for every ideal $I$ of $R$ that contains $J$, we have $I / J = \{ x + J \mid x ∈ I \} = \{ π(x) \mid x ∈ I \} = π(I)$.

How can I relate this to the correspondence isomorphism theorem for groups (which I assume it is related?)

It’s the same principle.
When working with groups, the canonical quotient homomorphism $π \colon G \to G/N$ induces bijections
\begin{align*}
  \{ \text{subgroups of $G$ that contain $N$} \}
  &⟷
  \{ \text{subgroups of $G/N$} \} \,,
  \\
  H &⟼ π(H) = H/N \,,
  \\
  π^{-1}(K) &⟻ K \,,
\end{align*}
and
\begin{align*}
  \{ \text{normal subgroups of $G$ that contain $N$} \}
  &⟷
  \{ \text{normal subgroups of $G/N$} \} \,,
  \\
  H &⟼ π(H) = H/N \,,
  \\
  π^{-1}(K) &⟻ K \,.
\end{align*}
When working with rings, we just copy-paste:
the canonical quotient homomorphism $π \colon R \to R/J$ induces bijections
\begin{align*}
  \{ \text{subrings of $R$ that contain $J$} \}
  &⟷
  \{ \text{subrings of $R/J$} \} \,,
  \\
  S &⟼ π(S) = S/J \,,
  \\
  π^{-1}(T) &⟻ T \,,
\end{align*}
and
\begin{align*}
  \{ \text{ideals of $R$ that contain $J$} \}
  &⟷
  \{ \text{ideals of $R/J$} \} \,,
  \\
  I &⟼ π(I) = I/J \,,
  \\
  π^{-1}(K) &⟻ K \,.
\end{align*}
(We have already encountered the last bijection at the beginning of this post.)

The notes say that computing $I/J$ corresponds to computing $φ(I)$ where $φ$ is the reduction homomorphism. Why is this the case?

Let us once again denote the canonical quotient homomorphism $R \to R/J$ by $π$.
Let us denote the constructed isomorphism from $R/J = ℤ[x] / J$ to $(ℤ/3ℤ)[x]$ by $ψ$.
The isomorphism $ψ$ comes from the first isomorphism theorem, and satisfies
$$
  ψ ∘ π = φ
$$
For every ideal $K$ of $R/J$, the subset $ψ(K)$ of $(ℤ / 3ℤ)[x]$ is again an ideal (because $ψ$ is an isomorphism), and the isomorphism $ψ$ induces an isomorphism
$$
  (R / J) / K ≅ ψ(R / J) / ψ(K) = (ℤ/3ℤ)[x] / ψ(K) \,.
$$
We are interested in the specific ideal $K = I / J$, for which we have
$$
  ψ(K) = ψ(I/J) = ψ(π(I)) = φ(I) \,.
$$
In other words:
together with the isomorphism $ψ \colon R / J ≅ (ℤ/3ℤ)[x]$ we get isomorphisms
$$
  R/I ≅ (R/J) / (I/J) ≅ (ℤ/3ℤ)[x] / ψ(I/J) = (ℤ/3ℤ)[x] / φ(I) \,.
$$
