What does it mean when I say that addition/multiplication for an equivalence relation is well defined? I have trouble understanding this concept. Why is it necessary to prove that addition or multiplication is well defined in equivalence classes? My understanding of equivalence classes is that it must be reflexive, symmetric and transitive. Doesn't proving it automatically imply that addition and multiplication can be done? Why the additional need to prove that it is 'well defined'?
Apologies if this question is too trivial; my understanding of this topic is limited.

Just because a relation is an equivalence, this doesn't mean it has to be "nice" with respect to any operation you'd like to put on its classes.  To see this, look at a non-example of something being well-defined. 
Let $V=\mathbb{R}^2$ be the plane with its usual vector space structure.   Put a relation on $V$ by defining $u \sim v$ if $u = cv$ for some non-zero scalar $c$.  This is an equivalence relation on $V$.  The zero vector is in a class by itself, and the other classes are the vectors that form parallel lines through the origin (with the zero vector removed).  
Since classes are naturally represented by vectors, you could attempt to define a natural addition on $V/\sim$ by 
$$
[u] + [v] = [u+v].
$$
But this fails to be well-defined, even though the relation in play is an equivalence.  
The problem is that two different people could add the SAME TWO classes but get different answers.  That's no good.  For example, Alice wants to add the class corresponding to the $x$-axis (with zero removed) to itself.  This class is $[(1,0)]$, for example. So, Alice does
$$[(1,0)] + [(1,0)] = [(2,0)] = [(1,0)].$$
Bob will do the exact same class addition, but recognize that $[(1,0)]$ is also equal to $[(-1,0)]$.  Then Bob gets
$$[(1,0)] + [(-1,0)] = [(0,0)] \neq [(1,0)].$$
Alice and Bob have added the exact same class to itself, but got conflicting answers. 
