Here is how I think about this; don't know if it will help.

Start with the straightening construction: this takes a map $f\colon X\to S$ of simplicial sets to a simplicial functor $\def\St{\mathrm{St}}\def\op{\mathrm{op}}\St(f)\colon \mathfrak{C}(S)^{\op}\to s\mathrm{Set}$.

The key example is when $f=\mathrm{id}\colon \Delta^n\to \Delta^n$. This is a simplicial functor $\St(\mathrm{id})\colon\mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$, which I'll call $\mathfrak{D}^n$. Recall that $\mathfrak{C}(\Delta^n)$ has mapping spaces $$\mathrm{map}(x,y) =N\bigl\{ \{x,y\}\subseteq S\subseteq [x,y] \bigr\},$$ i.e., nerves of posets of subsets of the interval $[x,y]$ containing the endpoints. The functor $\mathfrak{D}^n$ is defined similarly:
$$\mathfrak{D}^n(x) = N\bigl\{\{x\}\subseteq S\subseteq [x,n]\bigr\}.$$
The action $\mathfrak{D}^n(y)\times \mathrm{map}(x,y)\to \mathfrak{D}^n(x)$ is defined by union.

In general, we have
$$ \St(\Delta^n\xrightarrow{f} S) \approx \St(\Delta^n\xrightarrow{\mathrm{id}}\Delta^n)\otimes_{\mathfrak{C}(\Delta^n)} \mathfrak{C}(S),$$
where the tensor product really means left Kan extension along the (simplicial) functor $\mathfrak{C}(f)\colon \mathfrak{C}(\Delta^n)\to \mathfrak{C}(S)$ (or, more precisely, along its opposite).

The simplicial set
$$Q^n= \St(\Delta^n\to \Delta^0) \approx \mathfrak{D}^n\otimes_{\mathfrak{C}(\Delta^n)} 1,$$
where $1=\mathfrak{C}(\Delta^0)$ is the terminal simplicial category. In other words, the simplicial set $Q^n$ is the colimit of the (simplicial) functor $\mathfrak{D}^n\colon \mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$. This is just something you have to compute: it's a coequalizer of
$$\small
\coprod_{x=0,\dots,n} N\bigl\{ \{x\}\subseteq S\subseteq [x,n] \bigr\} \leftleftarrows \coprod_{0\leq x\leq y\leq n} N\bigl\{ \{y\}\subseteq S\subseteq [y,n]\bigr\} \times N\bigl\{ \{x,y\}\subseteq T \subseteq [x,y] \bigr\}.
$$

This turns out to be isomorphic to Lurie's quotient of $K_{[n]}$. Explicitly, a $k$-simplex in Lurie's quotient is a sequence $\varnothing \subsetneq S_0\subseteq \cdots\subseteq S_k\subseteq [n]$, subject to the relation that, for each $x\in S_0$, we identify it with the $k$-simplex $\varnothing \subsetneq S_0\cap[x,n]\subseteq \cdots \subseteq S_k\cap [x,n]\subseteq [n]$.