A connected covering is regular exactly when its induced subgroup is normal, exactly when deck transformations act transitively on a fibre

By [F2], every point of $p^{-1}(b_0)$ has the form $e_0\cdot g$ for some $g\in G$, and [L1] records the subgroup at that point.

By [L2], a deck transformation reaches $e_0\cdot g$ from $e_0$ exactly when $g$ normalizes $H$. Hence the deck action on $p^{-1}(b_0)$ is transitive exactly when $N_G(H)=G$, which by [F1] is exactly when $H⊴G$. This proves the equivalence of clauses 2 and 3.

For the implication from normality to regularity, clause 2 gives clause 3 by step 2.1. Let $e,e^{\mathrm{\prime }}$ lie over an arbitrary $b\in B$, choose a path from $b$ to $b_0$, and lift it from $e,e^{\mathrm{\prime }}$ to points $u,u^{\mathrm{\prime }}$ over $b_0$. Clause 3 gives a deck transformation $\tau$ with $\tau (u)=u^{\mathrm{\prime }}$. Applying $\tau$ to the reverse lift from $u$ produces a lift from $u^{\mathrm{\prime }}$, so uniqueness in [F3] gives $\tau (e)=e^{\mathrm{\prime }}$. Thus the deck group is transitive on every fibre and the covering is regular.

For the converse implication from regularity, the definition makes the deck action transitive on $p^{-1}(b_0)$, so clause 3 holds. Step 2.1 then gives $N_G(H)=G$, and [F1] gives $H⊴G$. Thus clauses 1, 2, and 3 are equivalent.