Let $H\leq S_n$ be a subgroup of order 12. Then $S_4/H$ is a group of order 2, hence $S_4/H \cong C_2$, which is abelian. Therefore, $\left[S_4,S_4\right] \leq H$, where $\left[S_4,S_4\right]$ denotes the commutator subgroup (smallest normal subgroup of $S_n$ with abelian quotient).
Now if we can show that $[S_4,S_4] = A_4$, we have $A_4 \leq H$, $\left|A_4\right| = \left|H\right|$ and therefore $A_4 = H$. Firstly, $S_4/A_4 \cong C_2$ as above, so $[S_4,S_4]\leq A_4$. For any 3-cycle $(i,j,k) \in S_4$:
\begin{align*}(i,j,k) = (i,k,j)^2 = ((i,k)(i,j))^2 &= (i,k)(i,j)(i,k)(i,j)\\ &= (i,k)(i,j)(i,k)^{-1}(i,j)^{-1} = \left[(i,j),(i,k)\right] \in \left[S_4,S_4\right].\end{align*}
The set of all 3-cycles in $S_4$ generate $A_4$, so $A_4 \leq \left[S_4,S_4\right]$ and the result follows.
If you want to show that $A_4$ really is a subgroup of index 2, define $\delta : S_n \rightarrow C_2$ by
$$\delta(\sigma) = \begin{cases} 1, &\sigma\; \text{even}\\
-1, &\sigma\; \text{odd}\end{cases}$$
and show it's an epimorphism.
To my mind it's cleaner to describe these things geometrically when possible.
$S_4$ naturally acts by rotations and reflections on a tetrahedron, permuting the four vertices. If we want to write down a map $S_4 \to S_3$ in a geometric way, we should find a set of $3$ things in a tetrahedron that rotations and reflections permute.
The clue is that a tetrahedron has $6$ edges: in fact the elements of $S_4$ naturally permute the set of pairs of opposite edges (all of which are visible in the above picture). This corresponds exactly to these partitions into pairs of subsets of size $2$, and hopefully everything else should be much clearer once you have this picture.
When Artin says permuting the indices permutes partitions, he means, for example, that if we have a permutation like $(123)$ that sends $1 \mapsto 2, 2 \mapsto 3, 3 \mapsto 1, 4 \mapsto 4$, it also acts on subsets of $\{ 1, 2, 3, 4 \}$ by permuting each individual element, so for example it sends the subset $\{ 1, 2 \}$ to the subset $\{ 2, 3 \}$, etc. and in this way it also permutes set partitions, e.g. sending the partition $12 | 34$ to the partition $23 | 14$. Again all of these can be visualized in terms of edges of the tetrahedron.
Best Answer
Let's prove injectivity first, since it is the easier claim.
Given some non-identity permutation $\sigma\in S_4$, consider $\sigma(\{1,2\}),\;\sigma(\{3,4\})$. If they aren't equal to $\{1,2\},\{3,4\}$ respectively, then $\sigma$ is mapped to a non-identity permutation as we desired. So assume that they are equal.
Then, either $\sigma(1)=2,\;\sigma(2)=1$ or $\sigma(3)=4,\;\sigma(4)=3$. Assume the former without loss of generality.
Then, $\sigma(3)\neq1$, so $1\notin\sigma(\{1,3\})$, which proves that $\sigma$ is not mapped to the identity permutation. This shows that the mapping is injective.
For the other part, show that the transposition that sends $1\to2$ and $2\to1$ maps to an even permutation. Thus, every transposition maps to an even permutation. Then, prove that the mapping respects composition, and you are done.
Proof of the mapping respecting composition:
Let $\sigma,\gamma\in S_4$, and let $F$ be the mapping in question.
For any $i\neq j\in\{1,2,3,4\}$, $$F(\sigma\circ\gamma)(\{i,j\})=\{\sigma(\gamma(i)),\sigma(\gamma(j))\}$$$$F(\sigma)\circ F(\gamma)(\{i,j\})=F(\sigma)(\{\gamma(i),\gamma(j)\})=\{\sigma(\gamma(i)),\sigma(\gamma(j))\}$$