Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$. That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$. Now if $X$ and $Y$ are orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.
So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.
On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,
$(X \star Y)^\mathrm{op} = X^\mathrm{op\cong Y^\mathrm{op} \star Y^\mathrm{op}.$X^\mathrm{op}.$So commuting$\star$distributes over$\mathrm{op}$: so in a sense, the only asymmetry in$\star$is an asymmetry of variance. This is nice and intuitive for ordered sets, and easily shown for all simplicial sets. 1 As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets. You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it should be commutative. Any order$X$can be seen as the simplicial set whose$n$-simplices are chains$(x_0 \leq \ldots \leq x_n)$from$X$. That is, there's a full and faithful "nerve" embedding$N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$. Now if$X$and$Y$are orders, seen as their nerves,$X \star Y$is exactly (the nreve of) their ordered disjoint union. So e.g.$1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$is an easy and intuitive example of the non-commutativity. On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets$X$,$Y$,$(X \star Y)^\mathrm{op} = X^\mathrm{op} \star Y^\mathrm{op}.$So commuting$\star$distributes over$\mathrm{op}$: so in a sense, the only asymmetry in$\star\$ is an asymmetry of variance. This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.