On the join of simplicial sets as a dependent product Prop. 3.5 of Joyal notes in quasicategories gives a description of $X\star Y$ as $i_*(X,Y)$, where $i^*\dashv i_*$ is the adjunction 
$$
i^*\colon \mathbf{sSet}/\Delta[1] \leftrightarrows \mathbf{sSet}/\partial\Delta[1]
$$
where $i^*$ is "pullback along $\partial\Delta[1]\to \Delta[1]$ and $i_*$ is the associated dependent product, which being $\partial\Delta[1]$ discrete should morally be a mere product; this baffles me a lot, since $X\star Y$ seems much more complicated than what I get if I blind myself and I try to write what $i_*(A,B)$ is for $A,B\in \bf sSet$ (confusing ${\bf sSet}/\Delta[1]$ with ${\bf sSet}^{\Delta[1]}$ it chooses $A\times B\to B$).
Here are the words of Joyal:

any clue?
I have some observations to make: somewhere there must be a typo, since either Joyal's symbol for adjointness is reversed, or he uses the left adjoint of pullback, $i_!$, which simply composes $A\to \partial\Delta[1]$ with $i$ (but this is pretty strange too, since then I am dealing with the mere coproduct map $X\sqcup Y\to \partial\Delta[1]\to\Delta[1]$). 
I have absolutely no clue about what he does in the proof so I tried something different: my alternative proof strategy was to simply prove that the unit $X\star Y\to i_*(X,Y)$ is a monic (only in this component); it's always a split epi since $i^*$ is a full functor. And, alas, I'm not able to prove this either!
 A: By the Yoneda lemma for the category $\mathbf{\Delta}/[1]$, using the canonical equivalence $(\mathbf{\Delta}/[1])^\wedge = \mathbf{\Delta}^\wedge/\Delta^1$, it suffices to show that the morphism in question induces a bijection
  $$ Hom((\Delta^n,f), X \star Y) \to Hom((\Delta^n,f), i_*(X \sqcup Y)) $$
for each $n$, and each $f : \Delta^n \to \Delta^1$.
By adjunction, it suffices to compute
  $$ Hom(i^*(\Delta^n,f), X \sqcup Y) $$
and compare the result with the formula in 3.2.
For each morphism $f$ (there are $n+2$ choices), it is easy to compute $i^*(\Delta^n,f)$ by hand; the possible values are listed in the formula Joyal writes.
A: This is coming way late, but I ran into similar trouble with this and figured I would expand on the answer given, as it is difficult for us beginners and there are never too many details to be shared.
We know that the pullback functor $i^*$ has a left and right adjoint, this is common to locally Cartesian closed categories, see this answer for more details. We typically denote the adjoints as $i_! \dashv i^* \dashv i_*$ or in the answer as $\Sigma_i \dashv i^* \dashv \Pi_i$. So yes Joyal did mess up some notation where he writes $i^* \vdash i_*$, but maybe that is just how he chooses his notation, idk. Now $i_!$ is easy to compute, it is basically post-composition, but by the answer I linked to, it is comparably difficult to compute $i_*$ directly. So in this solution we probe $i_*(X,Y)$ with simplices instead.
So suppose we have a morphism $\Delta^n \xrightarrow{f} \Delta^1$. We know that $\Delta^1 \cong \Delta^0 * \Delta^0$, and the nerve functor is fully faithful, so $\mathsf{sSet}(\Delta^n, \Delta^1) \cong \mathsf{Cat}([n], [1])$, so by the discussion in section 23.5 by Rezk, we know that $f$ decomposes uniquely into a map $\Delta^{i_f} * \Delta^{j_f} \xrightarrow{f_1 * f_2} \Delta^0 * \Delta^0$. Notice that the number of ways of decomposing $\Delta^n$ in this form is the same as the number of components in $$(X*Y)_n = X_n \sqcup X_{n-1} \times Y_0 \sqcup \dots \sqcup X_i \times Y_j \sqcup \dots \sqcup X_0 \times Y_{n-1} \sqcup Y_n$$
Now a morphism in $(\mathsf{sSet}/\Delta^1)(\Delta^n \xrightarrow{f} \Delta^1, i_*(X \sqcup Y) \xrightarrow{p} \Delta^1)$ is equivalent to a choice of simplex $x \in i_*(X \sqcup Y)$. However, by the adjunction this is also equivalent to a map in $(\mathsf{sSet}/\partial \Delta^1)(i^*(\Delta^n \xrightarrow{f} \Delta^1), X \sqcup Y)$, but by Lemma 3.4 in Joyal's notes linked above, $i^*(\Delta^n \xrightarrow{f} \Delta^0)$ is then the map $\Delta^{i_f} \sqcup \Delta^{j_f} \to \Delta^0 \sqcup \Delta^0$. Thus a map $i^*(\Delta^n \xrightarrow{f} \Delta^1) \to X \sqcup Y$ is just a choice of a $i_f$-simplex in $X$ and a $j_f$-simplex in $Y$, which we recognize as a choice of an $n$-simplex in $X * Y$. This provides us with our desired bijection. Since this bijection holds for all simplices, it tells us that there is an isomorphism $X * Y \cong i_*(X\sqcup Y)$.
