The bounded reachability recursion is correct

For $i=0$, [L1] says ${\mathrm{Reach}}_0^G(u,v)$ holds exactly when $u=v$ or there is an arc $u\to v$. These are precisely the directed paths from $u$ to $v$ of lengths $0$ and $1$, hence of length at most $2^0$.

Assume the claim for $i-1$ with $i\ge 1$. If ${\mathrm{Reach}}_i^G(u,v)$ holds, [L1] gives a midpoint $z$ with ${\mathrm{Reach}}_{i-1}^G(u,z)$ and ${\mathrm{Reach}}_{i-1}^G(z,v)$. By the induction hypothesis there are directed paths from $u$ to $z$ and from $z$ to $v$, each of length at most $2^{i-1}$, and concatenating them gives a path from $u$ to $v$ of length at most $2^i$.

Conversely, suppose there is a directed path $u=v_0\to v_1\to \cdots \to v_{\mathrm{\ell }}=v$ with $\mathrm{\ell }\le 2^i$. Put $m:=\min (2^{i-1},\mathrm{\ell })$ and choose $z:=v_m$. Then the initial segment from $u$ to $z$ has length at most $2^{i-1}$, and the remaining segment from $z$ to $v$ has length $\mathrm{\ell }-m\le 2^{i-1}$. The induction hypothesis gives ${\mathrm{Reach}}_{i-1}^G(u,z)$ and ${\mathrm{Reach}}_{i-1}^G(z,v)$, so [L1] yields ${\mathrm{Reach}}_i^G(u,v)$.

Steps 1.1, 1.2, and 1.3 prove the equivalence for every $i$.