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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?

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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.

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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 $S$.

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