= Solution
The relevant <saddle equilibrium> is the origin, with <eigenvalues>
$$
m_\pm=\frac{\kappa\pm\sqrt{\kappa^2-4\lambda}}2.
$$
For $\lambda<0$, they have opposite signs. Its <saddle index> is
$$
\boxed{\delta=-\frac{m_-}{m_+}
=\frac{\sqrt{\kappa^2-4\lambda}-\kappa}{\sqrt{\kappa^2-4\lambda}+\kappa}<1}
$$
near the homoclinic curve, because that curve has $\kappa>0$. Along its leading asymptote, $\delta=1-\tfrac45\sqrt{-\lambda}+o(\sqrt{-\lambda})$.
A local passage near the <saddle equilibrium> contributes a power $s^\delta$ to the <return map>. Because $\delta<1$, its <derivative> becomes large as the orbit approaches the <separatrix>. The <periodic orbits> meeting the loops are therefore unstable, and their periods diverge logarithmically. On the side between the homoclinic and subcritical Hopf curves there are two unstable single-well <periodic orbits>, surrounding the two <sink equilibria>. On crossing the double-loop curve downward, they join into one unstable outer <periodic orbit> surrounding both wells. The attracting outer <periodic orbit> is not the one that collides with the <saddle equilibrium>: it persists across this global event.
The unstable outer <periodic orbit> must subsequently meet that attracting outer <periodic orbit> in a fold of <periodic orbits>. This extra curve is required to obtain the six-region diagram. One can see its position from the same energy-balance calculation. After normalizing $a_0=1$, an outer Hamiltonian contour with $H>0$ selects
$$
\beta=B(H)=\frac{\int_0^{u_{\max}}u^2\sqrt{2H+u^2-u^4/2}\,du}
{\int_0^{u_{\max}}\sqrt{2H+u^2-u^4/2}\,du},
\qquad u_{\max}^2=1+\sqrt{1+4H}.
$$
At $H=0$, $B=4/5$. The <derivative> of the denominator diverges positively there, while that of the numerator stays finite, so $B$ initially decreases. Rescaling at large $H$ gives $B(H)\propto\sqrt H$, so it eventually increases. At its minimum the stable and unstable selected <periodic orbits> meet. Thus the <outer-cycle fold in a weakly perturbed double-well oscillator> has $\kappa_{\rm fold}=c_f(-\lambda)+o(|\lambda|)$ with $0<c_f<4/5$; quadrature gives $c_f\simeq0.75226$. The decreasing portion of $B(H)$ selects the unstable <periodic orbit>, and the increasing portion selects the stable one, since the <energy> drift has sign $\beta-B(H)$.
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