Homomorphism from $S_4$ to $S_3$
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.
Qiaochu Yuan gave you a nice intuition but let me give you a formal answer (after all the devil is always in details).
Let $\mathbb{N}_m=\{1,\ldots, m\}$ and let $\sigma\in S_m$. If $P=\{A_1,\ldots,A_k\}$ is a partitioning of $\mathbb{N}_m$ then
$$\sigma(P):=\{\sigma(A_1),\ldots, \sigma(A_k)\}$$
is a partitioning of $\mathbb{N}_m$ as well. You may think about $\sigma(P)$ as a "double image". So the idea here is to choose a collection of partitionings $\{P_1,\ldots, P_s\}$ in such a way that $\sigma(P_i)=P_j$ for any $\sigma\in S_m$. Because in this situation $S_m$ acts on $\{P_1,\ldots, P_s\}$.
In this concrete case we have $m=4$ and we deal with special partitionings: such that every element of the partitioning is a subset of size $2$. So how many such partitionings are there? Well, since we deal with a set of size $4$ then every such partitioning is uniquely determined by chosing a single subset of size $2$. And there are $6$ such subsets but each subset is paired with its complement which actually gives our three $\Pi_1,\Pi_2,\Pi_3$ partitionings. These are all there are under these constraints.
And since every permutation maps a subset of size $2$ to a subset of size $2$ and these are all possible partitionings into subsets of size $2$ then every permutation "double maps" each $\Pi_i$ to some $\Pi_j$. In other words $S_4$ acts on $\{\Pi_1,\Pi_2,\Pi_3\}$.
This induces a map $f:S_4\to S(\{\Pi_1,\Pi_2,\Pi_3\})$. Now when he says that $(1,2,3,4)$ acts on $\{\Pi_1,\Pi_2,\Pi_3\}$ as $(\Pi_1,\Pi_3)$ then it simply means that $f((1,2,3,4))=(\Pi_1,\Pi_3)$ or in other words that $(1,2,3,4)$ "double maps" $\Pi_1$ to $\Pi_3$. Note that the notation $(\Pi_1, \Pi_3)$ denotes the permutation $\tau:\{\Pi_1,\Pi_2,\Pi_3\}\to \{\Pi_1,\Pi_2,\Pi_3\}$ such that $\tau(\Pi_1)=\Pi_3$.