Showing the Function in Second Isomorphism Theorem is Well-Defined In particular this is done in the context of vector spaces (so I'm using abelian notation).
Let $S$ and $T$ be subspaces of $V$.  I am trying to show that $(S + T)/T \cong S/(S \cap T)$.
Then we define $\tau: S + T \rightarrow S/(S \cap T)$ s.t. $\tau(s + t) = s + (S \cap T)$.
I am trying to remind myself why $\tau$ is well-defined.
So let let $v \in S+T$ s.t. $v = s_1 + t_1 = s_2 + t_2$.  
Then $\tau(s_1 + t_1) = s_1 + (S \cap T)$
and $\tau(s_2 + t_2) = s_2 + (S \cap T)$
Then why must these two be equal?
 A: If you are doing the second isomorphism theorem, you can use the first one.
Then define a linear mapping
$$
S \to (S+T)/T
$$
by
$$
s \mapsto s+ T.
$$
(PS Actually, defining this mapping is best done in two steps, as in the answer of @ZevChonoles.)
Prove it is onto, and find its kernel, which will be $S \cap T$. The first isomorphism theorem gives you the isomorphism - in particular, it takes care of the problem whether the map is well defined.
In a sense, I suggest you to be lazy. You have already dealt with the well definition in the first isomorphism theorem. Now just use that result without going through the motions again.
A: You have that $s_1-s_2\in S$, of course, as well as
$$s_1-s_2=t_2-t_1\in T$$
so $s_1-s_2\in S\cap T$, and therefore $s_1+(S\cap T)=s_2+(S\cap T)$.
By the way, I find it to be easier to consider the map in the other direction, which is obtained as the composition
$$S\hookrightarrow S+T \twoheadrightarrow (S+T)/T.$$
Think about what the kernel of the composition is; then show that the composition is surjective because any element of $S+T$ is equivalent (modulo $T$) to one in $S$.
