Can we make sense, in general, of taking a quotient by multiple ideals? I feel that this is a rather silly question, stemming from a fundamental misunderstanding of quotients, but I'm not quite able to make it precise. My question is: given two ideals $\mathfrak{a,b}\subseteq R$ where $R$ is some commutative ring, can we in general give meaning to a quotient $(R/\mathfrak a)/\mathfrak b$?
For example, consider $R=k[x,y]$. The ring $(R/(x))/(y)\cong k[y]/(y)\cong k$ obviously makes sense.
Another example may be $R=\mathbb Z$ with ideals $(2),(4)$. Then, if we were to give some meaning to these quotients, then we might write $(\mathbb Z/(2))/(4)\cong\mathbb Z/(2)$, since in a sense when we take the second quotient we have already "cut out the relevant information about $\mathbb Z$" by quotienting by $(2)$.
My first question: does any of this make sense at all? In particular, can $(4)$ even be considered an ideal of $\mathbb Z/(2)$?
My second question: if the answer to the above is positive, and I have ideals $\mathfrak b\subseteq\mathfrak a\subseteq R$, then can I prove that $(R/\mathfrak a)/\mathfrak b\cong R/\mathfrak b$ generalizing the second example? If these ideas make sense, I would equivalently need to construct $\phi:R/\mathfrak a\to R/\mathfrak b$ with kernel $\mathfrak b$.
Finally, my motivation comes from exercise 1.8 in Hartshorne: in the solution, we have some affine variety $Y$ of dimension $r$ and an ideal $\mathfrak p$ corresponding to the intersection of $Y$ with a nontrivial hypersurface - we end up proving $\dim A(Y)/\mathfrak p = r-1$ and want to conclude that the variety associated to $\mathfrak p$ has dimension $r-1$. Thus, we'd need to show $\dim (A/I(Y))/\mathfrak p=\dim A(Y)/\mathfrak p$ which would follow from the second question, however all solutions that I have seen regard this as a somehow trivial step.
 A: Let $R$ be a commutative ring and consider three $R$-modules $A \subseteq B \subseteq C$. Then $B/A \subseteq C/A$ and we have the following

Proposition. $(C/A) \big/ (B/A) \simeq C/B$.
Proof. Let $\varphi \colon C/A \to C/B$ be the map defined (on the cosets) by $\varphi(x + A) = x + B$. Then $\varphi$ is a well defined module homomorphism with kernel $B/A$, so the statement holds thanks to the first isomorphism theorem.

In particular, an ideal is an $R$-module contained in $R$, so this applies to any chain of ideals $I \subseteq J \subseteq R$, too.
It is worth noting that we also have the following

Proposition. If $A,B$ are two submodules of an $R$-module $C$, then
  $$
(A+B)/A \simeq B/(A \cap B)
$$
  Proof. The composite module morphism $B \to A+B \to (A+B)/A$ is surjective and has kernel $A \cap B$, so the statement follows from the first isomorphism theorem.

Again, this holds in particular for any two ideals $I,J$ of $R$.
A: I think Darij Grinberg's comment deserves to be an answer. The answer by A.P. implicitly addresses your second question, but Grinberg's comment addresses both your questions, explicitly.
Yes, your ideas make sense. Here is explicitly how:
When you take a quotient of $R$ by an ideal $\mathfrak{a}$, you automatically get a homomorphism (the "canonical homomorphism") from $R$ to the quotient ring $R/\mathfrak{a}$. Call it $\pi: R\rightarrow R/\mathfrak{a}$.
The point is that $\pi$ maps $\mathfrak{b}$ to an ideal $\pi(\mathfrak{b})$ in $R/\mathfrak{a}$. This is what Michael Burr referred to in his comment. Darij wrote it as $\mathfrak{b}\cdot (R/\mathfrak{a})$. (This notation comes from regarding $R/\mathfrak{a}$ as an $R$-module.)
This ideal is called the "pushforward" of $\mathfrak{b}$. (Atiyah-Macdonald call it the "extension".) It is $\mathfrak{b}$'s natural counterpart in the ring $R/\mathfrak{a}$.
Thus you can now take the quotient ring of $R/\mathfrak{a}$ by this ideal $\pi(\mathfrak{b})$. This is how you take the quotient by two ideals.
Furthermore, by a not-deep theorem, the ring you get is isomorphic to if you had just taken the quotient of $R$ by the ideal $\mathfrak{a} + \mathfrak{b}$ in the first place:
$$\frac{R/\mathfrak{a}}{\pi(\mathfrak{b})} \cong \frac{R}{\mathfrak{a}+\mathfrak{b}}$$
This is essentially a consequence of A.P.'s Theorem 1, with $C=R$, $B = \mathfrak{a} + \mathfrak{b}$, and $A = \mathfrak{a}$. The only work is to see that $\pi(\mathfrak{b}) = (\mathfrak{a}+\mathfrak{b})/\mathfrak{a}$. One also needs to think a little bit about why the theorem (which is formulated in terms of modules) also gives a ring isomorphism if $C$ is a ring and $A,B$ are ideals.
The beautiful thing about this theorem is that it shows that you get the same thing whether you quotient out $\mathfrak{a}$ first or $\mathfrak{b}$, because $R/(\mathfrak{a}+\mathfrak{b})$ is symmetric in the two.
