Concept of an Equivalence class partition The definition:
Part 1:
The equivalence classes of an equivalence relation on a set S constitute a partition of S. 
Part2:
Conversely, for any partition of S, there is an equivalence relation on S whose equivalence classes are the element of p.
To make sure I'm not blundering over the definition, let's suppose
a and b are integers in the set S.
The equivalence class of $a$ is given as:
$$[a]=\{a+kn|k \in S\}$$
We have that:
$$[1]=\{\dots,(-1-kn),(-1-(k+1)n),\dots,(1+kn),(1+(k+1)n),\dots\}$$
and where a takes on discrete values, we have equivalence classes that forms a partition on a set S.
The partition is: $$[1],[2],[3],....$$
This concludes my understanding of the first part.
Is my understanding correct?
Could someone also shed some light on the second part?
 A: The partition claimed in your statement is correct iff the different $[i],[j]$ are all distinct.
I suggest you refrain from examples and remain on the mere abstract level:
The equivalence classes determine a partition of $S$ because any element of $S$ is contained in an equivalence class, and two equivalence classes are either disjoint or equal, due to the definition of equivalence relation.
On the other hand, given a partition, you can define an equivalence relation by setting any two elements in the same portion of the partition to be equivalent, and not equivalent if they lie in different component. You can easily verify that this defines an equivalence relation.
For the second paragraph: Assume your set $S$ admits a partition $S= \bigsqcup_{i\in I} P_i$. Then for any two elements $s_1,s_2\in S$ you define $s_1\sim s_2 :\iff \exists \ i\in I \ s.t. s_1,s_2\in P_i$. Now you can show that this defines an equivalence relation which conversely defines your partition you started with.
A: An example.
Take the integers, and consider this equivalence relation:
A is equivalent to b if they differ by three or a multiple of three.
So, we have
[1]={..., -5, -2, 1, 4, 7, ...}
[2]={..., -4, -1, 2, 5, 8, ...}
[3]={..., -3, 0, 3, 6, 9, ...}   
These are the only three classes of this relation.
Every integer appears in exactly one class.
Any element of a class can be taken as its representative ([3]=[0]).
Every equivalence relation works like this: it divides a set in "slices" that don't overlap, and cover the entire set. That is to say, a partition.  
Second part: if you already have a partition, you can find an equivalence relation with that partition. Just take something equivalent to "x and y are in the same slice"!
