Heuristically: Let us take a finite set $S= \{s_1,...,s_n\}$. Now we "glue" together some elements, for example we could glue together $s_1$ and $s_2$, if we consider them to be "same object" we obtain a new set $S^*= \{s_1=s_2, s_3,...,s_n\}$. How should we call this first object in $S^*$? We can call it by $s_1$ or $s_2$ because it is determined by one of the two, but it is clear that it is not the same object as $s_1$ or $s_2$ in $S$. What about calling it equivalence class of $s_1$ and $s_2$? Similarly we can call equivalence relation the glueing operation we perform on the set $S$.
Formally: This time the set $S$ is not finite anymore, simply because we don't need it. Let us cite the formal definition of equivalence relation from Wikipedia.
A given binary relation $\sim$ on a set $S$ is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. Equivalently, for all $a$, $b$ and $c$ in $A$: $ a \sim a. \qquad \text{(Reflexivity)}$ $\text{if} \: a \sim b \:\text{then} \:b \sim a. \qquad \text{(Symmetry)}$ $\text{if}\: a \sim b\:\text{and}\: b \sim c \:\text{then}\: a \sim c. \qquad \text{(Transitivity)}$
To translate what we said above now we glue the elements $s_1$ and $s_2$ of $S$ if and only if $s_1\sim s_2$. Why do we need this complicated definition? Because we can now make a partition of the set $S$, let us call $[s]=\{t\in S\; | \;t\sim s\}\subset S$ the set of all the elements in $S$ which are in relation with $s$, and make this construction for every element of the set.
It is clear that can happen $[s] \cap [t] \neq \emptyset$ for some elements $s,t \in S$, but using the definition of equivalence relation above you can easily verify that this is indeed a partition of the set $S$. Where this essentially means that if $[s]\cap [t] \neq \emptyset$ then $[s]=[t]$.
Now the last step is simply to give a name to the subset $[s]\subset S$, how do you like equivalence class of $s$?
Example: Let us examine your proposed example. Take $S=\mathbb{Z}$, the set of integers number. and the equivalence relation to be $a \sim b \Leftrightarrow a = 4k +b$ for some $k\in \mathbb{Z}$. Then you can use the machinery above for $s=0$: $[0]= \{n \in \mathbb{Z} \:|\: n= 4k \text{ for some } k \in \mathbb{Z}\}= \{...,-12,-8,-4,0,4,8,12,...\}.$ And for $s=1$: $[1]= \{n \in \mathbb{Z} \:|\: n= 4k +1 \text{ for some } k \in \mathbb{Z}\}= \{...,-13,-9,-5,-1,3,7,11,...\}.$