This is on page 67, Lem 23.14 of Rezk's notes.
I can intuitively see why this is true but cannot write this out neatly.
By computing component wise we may attempt to construct $f$ given a set of triplets.
$\pi:K_n \rightarrow \Delta^1_n$, sends an element $\alpha$ to its respective decomposition $(\Delta^0)_i \times (\Delta^0)_j$. This gives unique maps $[n] \rightarrow [i], [n] \rightarrow [j]$.
So we pullback the element $\alpha$ to $$(K_0)_i, (K_1)_j$$ and compose it via $f^0, f^1$.
$$ (K_0)_i \rightarrow X_i, (K_1)_j \rightarrow Y_j $$
sending to an element in $X_i \times Y_j \subseteq X \star Y$.
This would then be how I define $f:K \rightarrow X \star Y$.
This all seems very messy. I would appreciate if someone may elaborate the more general perspective and a neater proof.
Perhaps a way to make it cleaner is to break it up into two steps:
For any maps $g\colon A\rightarrow X$ and $h\colon B\rightarrow Y$, define $(g\star h)\colon A\star B\rightarrow X\star Y$.
Show that every map $K\rightarrow \{0\}\star\{1\}$ factors "uniquely" through $K^{\{0\}}\star K^{\{1\}}\rightarrow \{0\}\star\{1\}$ (where, to get true uniqueness, we have to require the restriction of the map $K\rightarrow K^{\{0\}}\star K^{\{1\}}$ to each $K^{\{i\}}$ to be the identity).
Then $f$ will be the composite of the induced map $K\rightarrow K^{\{0\}}\star K^{\{1\}}$ with $f_{\{0\}}\star f_{\{1\}}$.
There are still details to check here (and you really should check them, because I haven't and I could be wrong!), but maybe this presentation is a little more satisfying?