$k[x,y]/(x^2-y, y-r) = k[x]/(x^2-r)$ 
Show $R[x,y]/(x^2-y, y-r) = R[x]/(x^2-r)$

Intuitively it makes sense since the left hand side the quotient is just identifying $x^2$ as $y$ and $y$ as $r$.
To show this rigorously, I think I can use the isomorphism theorem
$$R[x,y]/(x^2-y, y-r) = \frac{R[x,y]/(y-r)}{(x^2-y, y-r)/(y-r)}$$
But then I know we can not replace things in the quotient with isomorphic things. For example $\mathbb{Z}/{2\mathbb Z} \neq \mathbb{Z}/{4\mathbb Z}$ where $2\mathbb{Z}$ and $4\mathbb{Z}$ are isomorphic as $\mathbb{Z}-module$.
I this case, can I say the top $R[x,y]/(y-r)$ is in fact just $R[x]$? And I think the botton is not even isomorphic (as Abelian group) to $J$ since $\frac{I+J}{I} = \frac{J}{I\cap J}$.
More generally, when can we replace things in a quotient by its isomorphic class.
 A: There is a unique $R$-algebra homomorphism from $R[x,y]$ to $R[x]/(x^2-r)$
taking $x$ to $x$ and $y$ to $r$. Its kernel contains both $x^2-y$ and $y-r$ so induces an $R$-algebra homomorphism from $R[x,y]/(x^2-y,y-r)$
to $R[x]/(x^2-r)$.
Similar considerations show there is an $R$-algebra homomorphism in
the reverse direction taking $x$ to $x$. All we need to prove is that
these maps are inverse. This requires seeing where $x$ and $y$ get sent
to in their composites.
A: I've decided to turn my comment into a proper answer.
When you are thinking of subobjects and quotient objects, you always have to have in mind that these actually come equipped with morhpisms. Subobject $A$ of $X$ comes with inclusion map $A\hookrightarrow X$ (or more generally, monomorphism) and quotient object with epimorphism $X\twoheadrightarrow X/A$. These are organized nicely with short exact sequence
$$\require{AMScd}
\begin{CD}
0 @>>> A @>i>> X@>p>> X/A@>>> 0
\end{CD}$$
and exactness means that $i$ is monomorphism, $p$ is epimorphism and $\operatorname{im}i=\ker p.$
What you want to know is if $X\cong Y$ and $A\cong B$, when is $X/A\cong Y/B$. Luckily, this can easily be answered using short exact sequences. 
Consider this situation:
$$\require{AMScd}
\begin{CD}
0 @>>> A @>i_A>> X @>p_A>> X/A @>>> 0\\
&     @VVV    @VVfV  @VV?V \\
0 @>>> B @>i_B>> Y @>p_B>> Y/B @>>> 0
\end{CD}$$
and the question is given morphism $f\colon X\to Y$ (not necessarily iso), when can we expect morphism $\bar f\colon X/A\to X/B$? Well, this is actually quite easy, we have $p_B\circ f\colon X\to Y/B$ and we know that we must have $A\subseteq\ker(p_B\circ f)$ which is the same thing as $f(A)\subseteq B$. So, we need to have well defined restriction $f|_A\colon A\to B$. This can be written as $f\circ i_A = i_B\circ f|_A$.
To summarize,

Lemma. Let $f\colon X\to Y$ and $g\colon A\to B$ such that $f\circ i_A = i_B\circ g$. Then there exists unique $h\colon X/A\to X/B$ such that $h\circ p_A = p_B \circ f$, i.e. there is following commutative diagram
  $$\require{AMScd}
\begin{CD}
0 @>>> A @>i_A>> X @>p_A>> X/A @>>> 0\\
&     @VVgV    @VVfV  @VV\exists!\, hV \\
0 @>>> B @>i_B>> Y @>p_B>> Y/B @>>> 0
\end{CD}$$

Written less mysteriously, $h([x]) =[f(x)]$, where $[\cdot]$ is appropriate equivalence class.
So, returning to the case of isomorphism $f\colon X\stackrel{\sim}\to Y$, we can have $X/A\cong Y/B$ only if $f(A) \subseteq B$. It turns out that $f(A) = B$ is enough, i.e.

Lemma. (addendum) If $f$ and $g$ above are isomorphisms, then so is $h$.

You can use known tools such as first isomorphism theorem to prove it or directly from five lemma.
As you can see, the problem with $2\mathbb Z$ and $4\mathbb Z$ is that even though they are isomorphic, there is no automorphism of $\mathbb Z$ (which is just $x\mapsto \pm x$) that restricts to it, and thus $\mathbb Z/2\mathbb Z\not\cong\mathbb Z/4\mathbb Z.$
Finally, let's see how to deal with your exercise using the tools we just mentioned. Write appropriate exact sequences:
$$\require{AMScd}
\begin{CD}
0 @>>> \frac{(x^2-y,y-r)}{(y-r)} @>>> \frac{R[x,y]}{(y-r)} @>>> \frac{R[x,y]/(y-r)}{(x^2-y,y-r)/(y-r)} @>>> 0\\
&     @VVV    @VVV  @VVV \\
0 @>>> (x^2-r) @>>> R[x] @>>> R[x]/(x^2-r) @>>> 0
\end{CD}$$
and note that isomorphism $R[x,y]/(y-r)\cong R[x]$ is given by evaluation map $$f(x,y) + (y-r)\mapsto f(x,r).$$ Now check that this evaluation map maps $(x^2-y,y-r)$ to $(x^2-r)$ and you are done.
Perhaps you've noticed that I haven't specified in which category is this isomorphism; take your pick: $\mathbb Z$-algebra, $R$-algebra or even $R[x]$-algebra - all you have to do is check that all the maps in the diagram are $R[x]$-linear and multiplicative (which all follows from evaluation map being such).
