Let $p:X \to S$ be a right fibration of simplicial sets; one can roughly think of it as some sort of "functor" $S^{op} \to Set_\Delta$ (where $Set_\Delta$ denotes simplicial sets) sending $s \in S$ to the fiber $X_s=p^{-1}(s)$ which is a Kan complex. Functor is in quotes because the problem is this isn't actual functor, i.e. for an edge $s \to s'$, you would need to define a map of simplicial sets $X_{s'} \to X_s$, but it is not clear how...you could also instead of trying to define a simplicial map use the correspondence $p^{-1}(s, s')$ but composition of correspondences gives some trouble. If I understand correctly (and I'm not sure I do, so please correct me), to construct an actual functor $F:\mathfrak{C}[S]^{op} \to Set_\Delta$ such that $F(s)$ is weak homotopy equivalent to $X_s$, Lurie in Higher Topos Theory 2.2.1 defines a straightening functor $St(X):\mathfrak{C}[S]^{op} \to Set_\Delta$ from the simplicial category $\mathfrak{C}[S]^{op}$ built out of $S$ to $Set_\Delta$ as follows: let $$\mathcal{M}=\mathfrak{C}[S] \underset{\mathfrak{C}[X]}{\coprod}\mathfrak{C}[X^\triangleright]$$ then for $s \in \mathfrak{C}[S]$, define $$St(X)(s)=Map_{\mathcal{M}}(s, \infty)$$ where $\infty$ is the cone point of $X^\triangleright$.
My question is why does this definition give what we want? In particular, is $Map_{\mathcal{M}}(s, \infty)$ weak homotopy equivalent to $X_s$? We also need $St X$ to be a simplicial functor and I suppose this comes about because $\mathcal{M}$ is a simplicial category, there is a simplicial map $$Map_{\mathcal{M}}(s, s') \times Map_{\mathcal{M}}(s', \infty) \to Map_{\mathcal{M}}(s, \infty)$$