Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.

For example, if S and S' are both constant simplicial sets {a} and {b} respectively, then the set of 1-simplices of the join (the value on {0,1}) has 3 elements: the element 'a' indexed by {0,1} cross the empty set; the element (a,b) indexed by {0} x {1}; and the element 'b' indexed by the empty set cross {0,1}. You're implicitly choosing a "direction" so that the lines in the join move from one simplicial set S to the other S'.

Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.

EDIT: Some elaboration.

For a simplicial set $T$, let's write $T\_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)

Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.

First, $(S \star S')\_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')\_0 = S\_0 \cup S'\_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.

Next, the 1-simplicies. Similarly $$ (S \star S')\_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S\_1 \cup (S\_0 \times S'\_0) \cup S'\_1 $$ There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.

The two boundary maps $(S \star S')\_1 \to (S \star S')\_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S\_0 \times S'\_0$ the back boundary is the projection to $S\_0$ and the front boundary is projection to $S'\_0$. There is asymmetry here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.