"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
Let us use the fat slice $\mathcal{C}^{z/}$ (See HTT, $\S$4.2.1) and the model $\operatorname{Hom}_{\mathcal{C}}(x,y)=\operatorname{Fun}(\Delta^1,\mathcal{C})\times _{\mathcal{C}\times \mathcal{C}}\{(x,y)\}$ of the mapping space. By computation, we can check that the square
$$\require{AMScd} \begin{CD} \operatorname{Hom}_{\mathcal{C}^{/z}}(f,g) @>>> \operatorname{Hom}_{\operatorname{Fun}(\Delta^1,\mathcal{C})}(f,g)\\ @VVV @VVV \\ \Delta^0 @>{\operatorname{id}_z}>> \operatorname{Hom}_{\mathcal{C}} (z,z)\end{CD}$$
is cartesian. The right vertical arrow is a Kan fibration, because $\operatorname{ev}_1:\operatorname{Fun}(\Delta^1,\mathcal{C})\to\mathcal{C}$ is an inner fibration. (In general, inner fibrations induce Kan fibrations between various mapping spaces. See, e.g., Lemma 2.4.4.1. The lemma talks about the usual slice, but the proof applies to fat slice as well.) So the above square is homotopy cartesian.