I've labeled signals in your block diagram.
Let's just walk back up the chain. There's a good chance I made some mistakes in my subscripts and delays, since I tried solving this on a post-it note... please feel free to edit if there are :D
y(k) &= v_3(k) \\
&= v_2(k-1) \\
&= v_1(k-2) \\
&= v_4(k-2) + v_5(k-2) + u(k-2) \\
Now, let's figure out what $v_5(k)$ and $v_4(k)$ are.
v_5(k) = v_3(k-1) = y(k-1)
v_5(k-2) = y(k-3).
Next, for $v_4(k)$,
v_4(k) &= -v_2(k) \\
&= -v_1(k-1) \\
This means that
$$ v_4(k-2) = v_4(k-3)-y(k-4)-u(k-3). $$
Plugging this all in,
$$y(k) = u(k-2)+y(k-3)+v_4(k-3)-y(k-4)-u(k-3)$$
This makes sense, since we write our observable state in terms of previous state observations $y(k-3),y(k-4)$, control input $u(k-2), u(k-3)$, and a single direct state measurement $v_4(k-3)$.
Edit: now the signal names match the diagram.