We describe the quotient rings in a concrete, computational way.
Let us review the definition, say for $\mathbb{Q}[x]/(x^2-2x)$. Two polynomials $A(x)$ and $B(x)$ are called equivalent modulo $x^2-2x$ if their difference is divisible by $x^2-2x$. Then $\mathbb{Q}[x]/(x^2-2x)$ officially consists of the equivalence classes, with the "natural" addition and multiplication modulo $x^2-2x$.
For any polynomial $A(x)$, there exist unique polynomials $Q(x)$ and $R(x)$ such that $R(x)$ has degree $\lt 2$ and
$$A(x)=(x^2-2x)Q(x)+R(x).$$
It follows that $A(x)$ is equivalent to $R(x)$. It is easy to see that two polynomials of degree $\lt 2$ are equivalent iff they are equal.
Thus the equivalence classes can be identified with the polynomials of degree $\lt 2$. Addition is the obvious one. Multiplication is a bit trickier. Let us see how to compute the polynomial of degree $\lt 2$ which is equivalent to $(ax+b)(cx+d)$.
Multiply in the natural way, to obtain $ac x^2+(ac+bd)x + bd$. Now recall that $x^2-2x$ is equivalent to $0$, so replace $x^2$ by $2x$ (to be explicit: $x^2 - 2x = 0 \Rightarrow x^2 = 2x$). We obtain
$(2ac +ac+bd)x +bd$. Now we have an explicit formula for the product.
An analogy: The ring $\mathbb{Z}/(12)$ is officially made up of equivalence classes, where $a$ and $b$ are called equivalent if their difference is divisible by $12$. But it is very useful to think of $\mathbb{Z}/(12)$ as made up of the objects $0,1,2, \dots, 11$, with new addition and multiplication. (For example, $5+9=2$, $5\times 9=9$.)
Similarly, the quotient rings we are looking at can be viewed either abstractly, or concretely as polynomials of degree $\lt 2$, with an unusual multiplication.
Since we have a concrete picture of what is going on, we should be able to answer some questions.
First we show that $\mathbb{Q}[x]/(x^2)$ is not isomorphic to $\mathbb{Q}[x]/(x^2-1)$.
Let's view $\mathbb{Q}[x]/(x^2)$ as the polynomials of degree $\lt 2$, with the natural multiplication except that $x^2$ is always replaced by $0$. Then, in the quotient ring, $(x)(x)=0$. (Strictly speaking, the equivalence class of $x$, times itself, is equal to the equivalence class of $0$.) So $\mathbb{Q}[x]/(x^2)$ has a non-zero element whose square is $0$.
We show that $\mathbb{Q}[x]/(x^2-1)$ has no such element. Suppose to the contrary that in $\mathbb{Q}[x]/(x^2-1)$, the square of (the equivalence class of)
$ax+b$ is $0$, that is, $(ax+b)^2$ is equivalent to $0$ modulo $x^2-1$.
Do the squaring. First we get $a^2x^2+2abx+b^2$. Then, since $x^2-1$ is equivalent to $0$, we replace $x^2$ by $1$, and obtain $2abx+b^2+1$. Could this be the $0$ polynomial? No, because the constant term $b^2+1$ cannot be $0$.
So $\mathbb{Q}[x]/(x^2)$ and $\mathbb{Q}[x]/(x^2-1)$ differ in a structural property: The first has a non-zero object whose square is $0$, and the second does not. But any isomorphism $\phi$ from $\mathbb{Q}[x]/(x^2)$ to $\mathbb{Q}[x]/(x^2-1)$ must preserve such structural properties. For completeness we do the details.
Suppose that $w\in \mathbb{Q}[x]/(x^2)$ is non-zero and $\phi(w^2)=0$, then $\phi(w)\ne 0$ and $0=\phi(w^2)=(\phi(w))^2$. So the square of $\phi(w)$ is $0$. This contradicts our earlier calculation, which showed that the square of a non-zero element of $\mathbb{Q}[x]/(x^2-1)$ cannot be zero.
Next we show that $\mathbb{Q}[x]/(x^2-1)$ is isomorphic to $\mathbb{Q}[x]/(x^2-2x)$.
Since $x^2-2x=(x-1)^2-1$, the equivalence class of $x$ in $\mathbb{Q}[x]/(x^2-1)$
should behave like the equivalence class of $x-1$ in $\mathbb{Q}[x]/(x^2-2x)$.
Since we are dealing with equivalence classes modulo two different polynomials, let us change to a more precise notation. Denote the equivalence class of $P(x)$ modulo $x^2-1$ by $P(x)/(x^2-1)$, and modulo $x^2-2x$ by $P(x)/(x^2-2x)$. (We should have used this more precise notation from the beginning, but avoided it for the sake of greater concreteness. But in what follows, don't let the $/(??)$ parts worry you, and maybe even omit them.)
So what should $(ax+b)/(x^2-1)$ be sent to by our isomorphism $\phi$? The natural choice is $(a(x-1) +b)/(x^2-2x)$.
It is clear that (equivalence classes modulo $x^2-1$ of) polynomials of degree $\lt 2$ are sent bijectively by $\phi$ to (equivalence classes modulo $x^2-2x$ of) polynomials of degree $\lt 2$. We must also check that $\phi$ preserves addition and multiplication.
Checking the addition is very easy. Let's deal with the multiplication. Look at $(ax+b)(cx+d)$. Modulo $x^2-1$, this is (equivalent to) $(ad+bc)x + ac+bd$.
Note that $\phi$ maps $(ax+b)/(x^2-1)$ to $(a(x-1)+b)/(x^2-2x)$ and maps
$(cx+d)/(x^2-1)$ to $(c(x-1)+d)/(x^2-2x)$. Also, $\phi$ maps
$((ad+bc)x + ac+bd)/(x^2-1)$ to $((ac+bd)(x-1)+ ac+bd)/(x^2-2x)$. So we need to verify verify that
$$(a(x-1)+b)(c(x-1)+d) \quad\text{is equivalent to}\quad (ac+bd)(x-1)+ ac+bd$$
modulo $x^2-2x$.
Multiply out the left-hand side, using the fact that $(x-1)^2-1$ is equivalent to $0$. We get $(ac+bd)(x-1)+ac+bd$, exactly what is wanted.
Comment: Look for example at $\mathbb{Q}[x]/(x^2-2x)$. The fact that $x^2-2x$ factors nicely means that we can express $\mathbb{Q}[x]/(x^2-1)$ as a direct product of simpler structures. This important structural information is easiest to approach through a more abstract approach. However, a concrete view of things is always useful, both for the understanding and for computational algebra.
Everything you wrote is correct and you have the right attitude : aggressively attacking simple examples and trying to see what is going on geometrically.
As for localization, you can use your last isomorphism if you want: localization does indeed commute with quotients.
But it is simpler to exploit your preceding isomorphism $ k[x,y]/\langle xy-1\rangle \cong k[x,x^{-1}]$ instead and to write
$$(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}\cong k[x,x^{-1}]_{{\langle x-1,x^{-1}-1\rangle}}=k[x,x^{-1}]_{{\langle x-1\rangle}}=k[x]_{{\langle x-1\rangle}} =k[x-1]_{{\langle x-1\rangle}}$$
Geometrical interpretation (very important!)
You are studying a hyperbola in the plane near the point $(1,1)$ by projecting it on the $x$-axis and you obtain an isomorphism with the affine line punctured at zero.
The projection sends $(1,1)$ to the point $x=1$ on the punctured line and the local ring $(k[x,y]/\langle xy-1\rangle)_{\langle x-1,y-1\rangle}$ of the hyperbola at $(1,1)$ isomorphically to the local ring of the line at $1$, namely $k[x-1]_{{\langle x-1\rangle}}$.
Note carefully that the puncture of the affine line plays no role in these local questions: the point $x=1$ only sees its immediate vicinity and doesn't care what happens at $x=0$.
Best Answer
To treat your special case $$\mathbb{Z}_{2}[X] / (x^4+1),$$ its elements are the (classes) of the remainders of Euclidean divisions by $x^4+1$, so $$\mathbb{Z}_{2}[X] / (x^4+1) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{2} \}.$$ (Here I am using brackets to denote residue classes.)
You sum the classes normally, and as to the product, you first take the product, and then take the remainder of Euclidean division by $x^4+1$. Alternatively, you take the product, and then use the relation $x^4 \equiv -1$ (which is the same as $x^4 \equiv 1$ in this case, as $1 = -1$ in $\mathbb{Z}_{2}$) to reduce the result. So it's not really different from computing in $\mathbb{Z}_{m}$, where you first take an ordinary sum or product, and then take the remainder modulo $m$.
So for instance $$ [1 + x^2] \cdot [1 + x^3] = [1 + x^2 + x^3 + x^5] = [1 + x + x^2 + x^3], $$ using one of the two methods above.
The general case of $F[x] / (f(x))$, with $F$ a field, is similar, you get the classes of the remainders of Euclidean division by $f(x)$.