Note that a homomorphism from $S_3$ to $\mathbb{Z}_6$ is a homomorphism into an abelian group. Therefore, there is a bijection
$\mathrm{hom}(S_3,\mathbb{Z}_6)\simeq\mathrm{hom}(S_3/[S_3,S_3],\mathbb{Z}_6),$
where $[S_3,S_3]$ is the normal subgroup of $S_3$ generated by the elements of the form $aba^{-1}b^{-1}$. Below it is demonstrated that $[S_3,S_3]=A_3$, so it follows from the fact that $A_3$ has index 2, that $\mathrm{hom}(S_3,\mathbb{Z}_6)\simeq\mathrm{hom}(\mathbb{Z}_2,\mathbb{Z}_6).$ There are only two elements in $\mathbb{Z}_6$ whose orders divide $2$, so it follows that there are only two homomorphisms from $S_3$ to $\mathbb{Z}_6$.
As Dylan pointed out in the comments, the subgroup $[G,G]$ of a group $G$ generated by the elements of the form $aba^{-1}b^{-1}$ is called the commutator subgroup of $G$. Note that each commutator of $S_n$ is an even permutation, and therefore $[S_n,S_n]$ is a subgroup of $A_n$. On the other hand, all the 3-cycles are commutators: $ (abc)=(ab)(ac)(ab)(ac). $ Since the subgroup $A_3$ is generated by the 3-cycles, it follows that $[S_3,S_3]=A_3$. It follows that $S_3/[S_3,S_3]=\mathbb{Z}_2$.
About the bijection: note that every homomorphism from an arbitrary group into an abelian group always sends the commutators to 0. In other words, the kernel of any such homomorphism always contains the commutator subgroup. It holds in more generality: If $G$ is a group and $A$ is an abelian group, then there is a bijection $ \mathrm{hom}(G,A)\simeq\mathrm{hom}(G/[G,G],A) $ More precisely, this bijection is given (from right to left) by precomposition with the quotient map $G\to G/[G,G]$. Thus, the bijection says the following: for every homomorphism $f:G\to A$ there exists a unique homomorphism $\tilde{f}:G/[G,G]\to A$ with the property that $f=\tilde{f}\circ\pi_{[G,G]}$. This is the universal property of the map (functor) $G\mapsto G/[G,G]$.