"This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$"
My previous answer was based on me misreading this quote :)
You want to show that this large diagram computes $\Sigma$. Lurie says some words about why that is : he says that the large diagram induces a morphism from one square to the next, namely from the square
\begin{CD}X @>>> 0 \\
@VVV @VVV \\
0 @>>> W\end{CD} to the square \begin{CD}Y @>>> 0 \\
@VVV @VVV \\
0 @>>> V\end{CD}
Now the first square (the second too, but I need it for the first) is left Kan extended from its restriction to the span $0\leftarrow X \to 0$ (that is what it means to be a pushout), so a map between these two squares is entirely determined by a map between the corresponding spans.
But now the second span (the first one too, but I need it for the second) is right Kan extended from the single vertex $Y$, so a map between the spans is the same thing as a map between these vertices. In this case, by design, the map between the vertices is $f$.
What I'm saying is : the map between the squares that you get from the large diagram, which one $W\to V$ is precisely your $u$ , is the only map (up to a contractible space) of squares which restricts to $f : X\to Y$. But the map of squares which induces $\Sigma f : \Sigma X\to \Sigma Y$ also restricts to $f$, by definition.
So the two maps of squares must be the same, and in particular the two maps $u$ and $\Sigma f: \Sigma X\to \Sigma Y$ must be the same.
So this reduces to showing that the two squares
(\begin{CD}X @>>> 0 \\
@VVV @VVV \\
0 @>>> W\end{CD} and the other one with $Y$)
are indeed pushout squares. But for both it follows from pasting of pushout squares.
EDIT : Regarding the comments. If you agree that the indexing category for the big diagram is a poset, then a map of squares in this diagram (i.e. a map $(\Delta^1)^3\to K$ where $K$ is the indexing category) is entirely determined by where it sends the vertices, together with the property that for each arrow in $(\Delta^1)^3$, the corresponding vertices in $K$ have an arrow between them.
So here, to get an arrow between the two squares in question from the big diagram, you need an arrow from $X$ to $Y$ (ok you have one, it's $f$), an arrow from $0$ to itself (ok, it's $id$), an arrow from $0'$ to $0''$ (ok, just go $0' \to Z\to 0''$), and finally an arrow from $W$ to $V$ (ok, it's $u$). Crucially, these arrows are in $K$, so you get a map $(\Delta^1)^3\to K$ (again, here I'm really using that $K$ is a poset) and thus, because your big diagram was originally something like $K\to \mathcal C$, you can precompose and get $(\Delta^1)^3\to \mathcal C$. Because of how you chose the vertices and arrows, this map of squares has, on the top left vertex $f$ and on the bottom right vertex $u$, therefore $u$ is identified with $\Sigma f$.
(note : I said that I used that $K$ was a poset, but in fact if $K$ were only a $1$-category, all hope would not have been lost, as we would simply have had to further check that some diagrams commute - in a poset, any diagram commutes so we can skip this step).
(another way to phrase this is that (nerves of) posets are $1$-coskeletal )
EDIT 2 : I was a bit quick when I said $K$ is a poset. I meant that the $\infty$-category presented by $K$ was a poset, i.e. that $K$ was categorically equivalent to a poset. You can see this by computing $\mathfrak C[K]$, which you can do by explicitly writing $K= (\Delta^1)^2\coprod_{\Delta^1}(\Delta^1)^2\coprod_{\Delta^1}(\Delta^1)^2$ (which you can easily simplify, up to categorical equivalence to $(\Delta^1\times \Delta^2)\coprod_{\Delta^{\{1\}}\times\Delta^{\{1,2\}}} (\Delta^1 \times \Delta^1)$, and then you have to do a bit of work)
Best Answer
I don't believe the first claim is true, but I can give a somewhat formal argument for the connectivity of $\Delta_{\le n} \times_{\Delta} \Delta_{/m}$. Let us write $u$ for the inclusion $\Delta_{\le n} \to \Delta$. The crucial result follows the following special case of Appendix A of https://arxiv.org/abs/2207.09256:
Theorem 1. Given a diagram $F : \Delta \to \mathcal{S}_{\le n-1}$ the canonical map $\lim_{\Delta} F \to \lim_{\Delta_{\le n}} u^*F$ is invertible.
(This may already be the result you were after, but we can carry this through to actually obtain the connectivity result you requested.)
Fix an arbitrary $F : \Delta \to \mathcal{S}_{\le n-1}$. By Theorem 1, we know that the following is an equivalence:
\begin{align*} &\lim F \\ &\cong \hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F) \\ &\to \hom_{\mathbf{Fun}(\Delta_{\le n},\mathcal{S})}(u^*\mathbf{1}, u^*F) \\ &\cong \hom_{\mathbf{Fun}(\Delta_{\le n},\mathcal{S})}(\mathbf{1}, u^*F) \\ &\cong \lim u^*F \end{align*}
Transposing, we conclude that $\hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F) \to \hom_{\mathbf{Fun}(\Delta,\mathcal{S})}(u_!u^*\mathbf{1}, F)$ is an equivalence and, consequently, that $u_!u^* \mathbf{1} \to \mathbf{1}$ is orthogonal to any $(n-1)$-truncated object in $\mathbf{Fun}(\Delta, \mathcal{S})$. Accordingly, $u_!u^*\mathbf{1} = u_!\mathbf{1}$ is $n$-connected. We therefore conclude that $(u_!\mathbf{1})(m) : \mathcal{S}$ is $n$-connected for any $m : \Delta$. Let us unfold this using the formula for left Kan extensions: $$ (u_!\mathbf{1})(m) = \mathrm{colim}_{ \Delta_{\le n} \times_{\Delta} \Delta_{/m} } \mathbf{1} = L(\Delta_{\le n} \times_{\Delta} \Delta_{/m}) $$
The conclusion then follows.
Incidentally, this is just slight alteration of the usual proof of one direction of Quillen's theorem A. The other direction (that suitably connective fibers implies $n$-cofinality) also holds.