The bounded reachability recursion uses logarithmic recursion depth

If $N=1$, the only possible path question is settled by ${\mathrm{Reach}}_0^G$. Suppose now that $N\ge 2$. Any directed path with repeated vertices can be shortened by deleting a directed cycle, so if $u$ reaches $v$ at all then $u$ reaches $v$ by a simple directed path. A simple directed path in an $N$-vertex digraph has length at most $N-1$, and $2^{\lceil {\log }_2(N-1)\rceil }\ge N-1$.

By [L1], every recursive call decreases the index $i$ by one, so starting from that initial $i$ yields recursion depth at most $i+1=O(\log N)$.

One stack frame stores the current vertices $u$, $v$, a midpoint candidate $z$, and the current index $i$. Each of these numbers is below $\max (N,i+1)$, so one frame uses $O(\log N)$ bits. Multiplying by the depth from step 2.1 gives total space $O({\log }^{2}N)$.

If $N\le 2^{cs(n)}$, then $\log N\le cs(n)$, and step 3.1 becomes $O(s(n{)}^2)$. The last sentence follows; [L2] records exactly the configuration bound that supplies such an $N$ in Savitch's theorem.