An ideal and all elements of the quotient ring 
 Defintion 1  An ideal I of R is an additive subgroup so that $$a\in R , s\in I \Rightarrow as,sa \in I$$
The ring R/I is called the quotient ring.
 Example 1  : $R=\mathbb{Z}[x], I=nR (n\in \mathbb{Z})$. Then $R/I "=" (\mathbb{Z}/n\mathbb{Z})[x]$. Or $I=xR$. What are the $A+I (A\in \mathbb{Z}[x])?$
A=a+bx+... , so they are equal to the $a+I (a\in \mathbb{Z})$ because if $A=a+bx+...$, then $A-a = bx+... \in I$. So $R/I "=" \mathbb{Z}$ or : $I=2R+xR = \{ 2A+xB; A,B \in R\}$. So we can say that there are at most $2$ different side classes $0+I, 1+I$ and they can not be equal because otherwise $1\in I$, so $1=2A+xB; A,B\in \mathbb{Z}[x]$. It is true that $R/I$ is identical to $\mathbb{F}_{2}=\{0,1\}$
 Example 2:  R=$\mathbb{F}_{2}[x]$, I=xR, $R/I = \{0+I, 1+I\} "=" \mathbb{F}_{2}$. Or $I=x^{2}R, A+I = (a+bx+...)+I = (a+bx)+I (a,b \in \mathbb{F}_{2})$. So there are at most 4 side classes : $0+I,1+I,x+I,x+1+I$ and they are all different from eachother.

This is an outtake from my script. I do not understand how in example 1 it is concluded that there are at most 2 different side classes and 4 side classes with $(\mathbb{F}_{2}/x^{2}\mathbb{F}_{2})[x]$ in the example 2. How does one find the side classes? Are they elements of the ring? It seems to be easy to find them for simple rings (the elements of $\mathbb{F}_{5}$ are the equivalence classes 0,1,2,3,4) , but how to find them (the elements) for quotient rings?

"Example 1" seems to be three examples: (i) $R/I$ with $R=\mathbb{Z}[x]$, $I=nR$; (ii) $R/I$ with $I=xR$; and (iii) $R/I$ with $I=2R+xR$. 
In the third case, $I$ consists of all polynomials with integer coefficients that have even constant term: all such polynomials can be written as a multiple of $2$ plus a multiple of $x$; and conversely, if a polynomial is a multiple of $2$ plus a multiple of $x$, then its constant term must be even.
When are $a_0 + a_1x+\cdots+a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ in the same lateral class modulo this $I$? If and only if
$$(a_0+\cdots + a_nx^n) - (b_0 + \cdots + b_mx^m) = (a_0-b_0) + (a_1-b_1)x + \cdots \in I.$$
In order for the difference to be in $I$ what do we need? We need $a_0-b_0$ to be even; so the two polynomials are in the same coset if $a_0$ and $b_0$ have the same parity. So there are at most two lateral classes: one for polynomials with even constant term, one for polynomials with odd constant term.
(Conversely, polynomials with constant terms of opposite parity are not in the same lateral class, so there are in fact exactly two lateral classes).
In Example 2, you have $R=F_2[x]$ and $I=x^2R$. A polynomial is in $I$ if and only if it is a multiple of $x^2$. If $a_0+a_1x+\cdots + a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ have $a_0=b_0$ and $a_1=b_1$, then their difference is a multiple of $x^2$; since there are two possible choices for $a_0$ and two possible choices for $a_1$, there are at most four possible lateral classes: one for each choice combination. Because every polynomial will have constant and linear terms equal to $0$, both equal to $1$, constant term $0$ and linear term $1$, or constant term $1$ and linear term $0$. So there are at most 4 possibilities that yield different lateral classes.
(Conversely, those four possibilities are pairwise distinct, since the difference of, say, a poynomial with constant term 1 and linear term 0, and one of linear and constant terms equal to $1$, would not be a multiple of $x^2$).
