Let $K$ be the subgroup of $S_4$ such that $K = \{e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$.
I'm not sure if the first part of the question is related, but it asks to show that $K \triangleleft S_4$.
I believe I've shown that $K \triangleleft S_4$.
For any $g \in S_4$, $g$ can be expressed as a product of transpositions
The definition of normal means we need to show $gKg^-1 = K$
So suppose we represent $K$ as arbitrary transpositions then: $K = \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\}$
Then for an arbitrary transposition in $S_4$:
$$(a,b) \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\} (a,b)$$ $$= \{e, (a,b)(c,d), (a,c)(b,d), (a,d)(b,c)\}$$ $$ = K$$
So $K$ is normal.
Now I need to show that $S_4/K\cong S_3$.
I'm pretty sure that I have to find a map $\phi:S_4 \rightarrow S_3$ that is a homomorphism such that $\ker(\phi) = K$, but I'm not sure how to do this.
Your proof that $K$ is a normal subgroup of $S_4$ is not quite sufficient. It is enough to show that $\tau K\tau=K$ for any transposition $\tau\in S_4,$ but you haven't justified it. (Of course, you don't have to, if you already know that result.) So, take an arbitrary transposition $(a,b)\in S_4$ and show that $$(a,b)\bigl\{e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\bigr\}(a,b)=K,$$ which you still haven't done.
Now, you could then finish the proof that $S_4/K\cong S_3$ by finding a homomorphism as you describe. Alternately, note that $S_4/K$ is a group (why?) with $6$ elements (why?), so is either cyclic or isomorphic to $S_3$ (why?).