I have trouble understanding how to multiply elements in $S_3$ with elements in $\mathbb{Z}_2$. I believe I am supposed to use that if $\circ$ is the operation on $G$ and $\bullet$ is the operation on $H$, then $(g,h)*(g',h') = (g\circ g', h \bullet h')$.
So the group operation for $S_3$ is composition. The group operation for $\mathbb{Z}_2$ is multiplication modulo $2$. If $g$ is the function that takes $123$ to $213$ and $h$ is the function that takes $123$ to $321$, and lets say $g' = 0$ and $h' = 1$, then we'd have $(g,h)*(0,1) = (g \circ 0, h \bullet 1), $ but that doesn't really seem to make sense, at least not in a way that I understand.
Can I perhaps choose some other group operation for $\mathbb{Z}_2$? Any suggestions?
Context: I am trying to understand why/how $S_3 \times \mathbb{Z}_2$ is a non-abelian group of order $12$, and particularly what the elements in this group look like.
What you wrote is the group composition for the direct product $S_3\times \Bbb Z/2$, but you applied it in the wrong order, i.e., the pair $(0,1)$ is not in $S_3\times \Bbb Z/2$, as it should.
The group $S_3\times \Bbb Z/2$ consists of the following $12$ elements: $$ (id,0),((123),0),((132),0),((12),0),((13),0),((23),0)), (id,1),((123),1),((132),1),((12),1),((13),1),((23),1)). $$
This is how a direct product $G\times H$ of two groups is defined as a set. This set of ordered pairs becomes a group via the composition $$ (g,h)\circ (g',h')=(gg',hh'). $$