Every element of $G$ can be written uniquely as $x^ay^b$ with $0\leq a\lt 4$ and $0\leq b\lt 4$, giving you the 16 elements of $G$. Multiplication is by addition of exponents, modulo $4$.
Of these, the only elements in $\langle x^2y^2\rangle$ are $x^0y^0 = e$, and $x^2y^2$ (since $(x^2y^2)^2 = x^4y^4 = x^0y^0$). Let's call this subgroup $N$.
What are the elements of $G/N$? They are congruence classes modulo $N$. When are two elements of $G$ congruent modulo $N$? $x^ay^bN = x^ry^sN$ if and only if eiether $x^ay^b=x^ry^s$, or $x^ay^b = (x^ry^s)(x^2y^2)$.
So, for example, $x^1y^0N = x^3y^2N$; $x^2y^0N = x^0y^2N$; $x^3y^0N = x^1y^2N$; and so on.
So what is important is to keep track of which element you multiply by $x^2y^2$ and what element you get.
In the end, what are the eight different cosets? You can take any element of $G$ and get a coset: and if you add two to the exponents of $x$ and $y$ (modulo $4$), you get the other "name" for the coset. So we have:
- $\overline{x}^0\overline{y}^0 = x^0y^0N = x^2y^2N = \overline{x}^2\overline{y}^2$.
- $\overline{x}^1\overline{y}^0 = x^1y^0N = x^3y^2N = \overline{x}^3\overline{y}^2$.
- $\overline{x}^2\overline{y}^0 = x^2y^0N = x^0y^2N = \overline{x}^0\overline{y}^2$.
- $\overline{x}^3\overline{y}^0 = x^3y^0N = x^1y^2N = \overline{x}^1\overline{y}^2$.
- $\overline{x}^0\overline{y}^1 = x^0y^1N = x^2y^3N = \overline{x}^2\overline{y}^3$.
- $\overline{x}^1\overline{y}^1 = x^1y^1N = x^3y^3N = \overline{x}^3\overline{y}^3$.
- $\overline{x}^2\overline{y}^1 = x^2y^1N = x^0y^3N = \overline{x}^0\overline{y}^3$.
- $\overline{x}^3\overline{y}^1 = x^3y^1N = x^1y^3N = \overline{x}^1\overline{y}^3$.
$Z_4\times Z_2$ is the set of all pairs of the form $(a,b)$ with $a\in Z_4$ and $b\in Z_2$.
Remember that isomorphism does not mean they are identical, it just means that there is a way of identifying the two groups so that the operations "match up". So it doesn't matter that $G/N$ has "one component" and $Z_4\times Z_2$ has "two components"; that's just how we are writing them, not what group they are.
To give you a hint about what the identification might be, you'll want to define a function that maps an element of the form $\overline{x}^a\overline{y}^b$ in $G/N$ to an element of the form $(r,s)$ in $Z_4\times Z_2$. This identification must be such that if $\overline{x}^a\overline{y}^b$ maps to $(r,s)$, and $\overline{x}^c\overline{y}^d$ maps to $(t,u)$, then the product $\overline{x}^{a+c\bmod 4}\overline{y}^{b+s\bmod 4}$ must map to $(r,s)+(t,u) = (r+t,s+u)$.
As a further hint, look at the first column of my list above. Notice that the exponent of $\overline{x}$ is always between $0$ and $3$ (just like elements of $Z_4$) and that the exponent of $y$ is always either $0$ or $1$ (just like the elements of $Z_2$). Might the obvious thing to try work?
Now, the above is fine, but it doesn't give much intuition about what is "going on" (and I apologize for that... my only excuse is that it was late and I was tired).
It should be clear that elements of $G$ can be written as $x^ay^b$ with $0\leq a,b\lt 4$, with the group operation being "add exponents modulo $4$", $x^ay^b\cdot x^ry^s = x^{a+r\bmod 4} y^{r+s\bmod 4}.$ What does taking the quotient modulo $\langle x^2y^2\rangle$ do?
You can think of taking the quotient modulo $x^2y^2$ as "forcing $x^2y^2=1$ and seeing what you get". If you make $x^2y^2=1$, then you make $y^2 = (x^2)^{-1}=x^2$. That means that any time you have an element of the form $\overline{x}^a\overline{y}^b$, if you have $b\geq 2$, you can replace $\overline{y}^2$ with $\overline{x}^2$ and still have the same element.
So this immediately tells you that you can write each element of $G/\langle x^2y^2\rangle$ as $\overline{x}^a\overline{y}^b$, and restrict $b$ to $b\in\{0,1\}$: because if you had a square, you can replace it with an $x^2$ instead, and if you had $\overline{y}^3$, you can replace it with $\overline{x}^2\overline{y}$. In essence, you never need more than one $\overline{y}$ for any element. So you can certainly "cut down" on the $\overline{y}$s.
However, notice that the "obvious thing" I suggest you try first above the line does not actually work: it doesn't work because if you map $\overline{x}^0\overline{y}^1$ to $(0,1)$ in $Z_4\times Z_2$, then you would need its square to be trivial; but $\overline{x}^0\overline{y}^1\cdot\overline{x}^0\overline{y}^1 = \overline{x}^0\overline{y}^2 = \overline{x}^2\overline{y}^0$ which is not trivial. Instead we need to be a bit more careful; mapping $\overline{x}^1\overline{y}^0$ to $(1,0)$ is okay as far as powers of $\overline{x}$ go. Now, $(2,0)$ in $Z_4\times Z_2$ is twice $(1,0)$, but it is also equal to twice $(1,1)$, twice $(3,0)$, and twice $(3,1)$. Since $\overline{x}^2\overline{y}^0$ is the square of $\overline{x}^1\overline{y}^0$ and $\overline{x}^0\overline{y}^1$, $\overline{x}^3\overline{y}^0$, and $\overline{x}^2\overline{y}^1$, this suggests that if we map $\overline{x}^1\overline{y}^0$ to $(1,0)$, we should map $\overline{x}^1\overline{y}^1$ to $(0,1)$, since then we get $\overline{x}^2\overline{y}^1$ corresponding to $(1,1)$, $\overline{x}^3\overline{y}^0$ corresponding to $(3,0)$, and $\overline{x}^0\overline{y}^1$ to $(3,1)$.
This makes more sense with the realization that $\overline{y}^2$ is not trivial, but rather equal to $\overline{x}^2$.