consists of all vertices of $G$ that, in $H$, lie in the strict exterior of $S_u$ and in the strict interior of $S_v$ simultaneously, or lie in $S_v \cup S_v$. In particular, any cycle of $G_{u,v}$ is also a cycle in $G$.
Now, $\mathcal{C}(u, v)$ is computed as follows. Start with an empty sequence, and the graph $G_{u,v} - (S_u \cup S_v)$. As long as there is a cycle in the current graph such that all vertices of $S_u$ are in the strict interior of $C$ with respect to $H$, remove vertices of degree at most 1 in the current graph until no such vertices remain, and append the outer face of the current graph as a cycle to the constructed sequence. It is clear that this process terminates in linear time, and that by the above discussion, it constructs a sequence of concentric cycles $\mathcal{C}(u, v) = (C_1, C_2, \dots, C_p)$ in $G$ such that $S_u$ is in the strict interior of $C_1$ in $H$, $S_v$ is in the strict exterior of $C_p$ in $H$.
To assert the existence of a path $\eta$ in $H$ with one endpoint $v_0 \in S_u$ and the other endpoint $v_{p+1} \in S_v$, such that the intersection of $V(\eta)$ with $V(G) \cup S_u \cup S_v$ is ${v_0, v_1, \dots, v_{p+1}}$ for some $v_i \in V(C_i)$ for every $i \in {1, \dots, p}$, we require the following claim.
Claim 6.1. Let $C_{p+1} = H[S_v]$. For every $i \in {1, 2, \dots, p}$ and every vertex $w \in V(C_i)$, there exists a vertex $w' \in V(C_{i+1})$ such that $w$ and $w'$ lie on a common face in $G_{u,v}^+$. Moreover, there exist vertices $w \in S_u$ and $w' \in V(C_1)$ that lie on a common face in $G_{u,v}^+$.
Proof of Claim 6.1. Consider $i \in {1, 2, \dots, p}$ and a vertex $w \in V(C_i)$. We claim that $\mathrm{rdist}(w, V(C_{i+1})) \le 1$, i.e. there must be $w' \in V(C_{i+1})$ such that $w, w'$ have a common face in $G_{u,v}^+$. By way of contradiction, suppose that $\mathrm{rdist}(w, V(C_{i+1})) \ge 2$. Then, by Proposition 6.1, there is a cycle $C$ that separates $w$ and $V(C_{i+1})$ in $G_{u,v}^+$ such that $\mathrm{rdist}(w, w'') = 1$ for every vertex $w'' \in V(C)$. Here, $w$ lies in the strict interior of $C$, and $C$ lies in the strict interior of $C_{i+1}$. Further, $C$ is vertex disjoint from $C_{i+1}$, since $\mathrm{rdist}(w, w') \ge 2$ for every $w' \in V(C_{i+1})$. Now, consider the outer face of $G[V(C_i) \cup V(C)]$. By the construction of $C_i$, this outer face must be $C_i$. However, $w \in V(C_i)$ cannot belong to it, hence we reach a contradiction.
For the second part, we claim that $\mathrm{rdist}(S_u, V(C_1)) \le 1$, i.e. there must be $w \in S_u$ and $w' \in V(C_1)$ such that $w, w'$ have a common face in $G_{u,v}^+$. By way of contradiction, suppose that $\mathrm{rdist}(S_u, V(C_1)) \ge 2$. Then, by Proposition 6.1, there is a cycle $C$ that separates $S_u$ and $V(C_1)$ in $G_{u,v}^+$ such that $\mathrm{rdist}(S_u, w'') = 1$ for every vertex $w'' \in V(C)$. Further, $C$ is vertex disjoint from $C_1$, since $\mathrm{rdist}(S_u, w') \ge 2$ for every $w' \in V(C_1)$. However, this is a contradiction to the termination condition of the construction of $\mathcal{C}(u,v). ◦$
Having this claim, we construct $\eta$ as follows. Pick vertices $v_0 \in S_u$ and $v_1 \in V(C_1)$ that lie on a common face in $G_{u,v}^+$. Then, for every $i \in {2, \dots, p+1}$, pick a vertex $v_i \in V(C_i)$ such that $v_{i-1}$ and $v_i$ lie on a common face in $G_{u,v}^+$. Thus, for every $i \in {0, 1, \dots, p}$, we have that $v_i$ and $v_{i+1}$ are either adjacent in $H$ or there exists a vertex $u_i \in V(H) \setminus V(G)$ such that $u_i$ is adjacent to both $v_i$ and $v_{i+1}$. Because $\mathcal{C}(u,v) = (C_1, C_2, \dots, C_p)$ is a sequence of concentric cycles in $G$ such that $S_u$ is in the strict interior of $C_1$ and $S_v$ is in the strict exterior of $C_p$, the $u_i$'s are distinct. Thus, $\eta = v_0 - u_0 - v_1 - u_1 - v_2 - u_2 - \dots - v_p - u_p - v_{p+1}$, where undefined $u_i$'s are dropped, is a path as required.
Finally, we argue that $p \ge 100 \cdot \alpha_{\mathrm{sep}}(k)$. Note that $100\alpha_{\mathrm{sep}}(k) = 100(\frac{7}{2} \cdot 2^{ck} + 2) \le 400 \cdot 2^{ck}$, thus it suffices to show that $p \ge 400 \cdot 2^{ck}$. To this end, we obtain a lower bound on the radial distance between $S_u$ and $S_v$ in $G_{u,v}^+$. Recall that $|S_u|, |S_v| \le \alpha_{\mathrm{sep}}(k) = \frac{7}{2} \cdot 2^{ck} + 2 \le 4 \cdot 2^{ck}$. Let $P = \mathrm{path}{R^2}(u,v)$, and recall that its length is at least $\alpha{\mathrm{long}}(k) = 10^4 2^{ck}$. Since $R^2$ has not detour, $P$ is a shortest path in $H$ between $u$ and $v$, thus for any two vertices in $V(P)$, the subpath of $P$ between them is a shortest path between them. Now, recall the vertices $u' = u'P$, $v' = v'P$ (defined in Lemma 6.5), and denote the subpath between them by $P'$. By construction, $|E(P')| \ge \alpha{\mathrm{long}}(k) - 2 \cdot \alpha{\mathrm{pat}}(k) = (10^4 - 200) \cdot 2^{ck}$.
We claim that the radial distance between $S_u$ and $S_v$ in $G_{u,v}^+$ is at least $|E(P')|/2 - |S_u| - |S_v|$. Suppose not, and consider a sequence of vertices in $G_{u,v}^+$ that witnesses this fact: $x_1, x_2, x_3, \dots, x_{p-1}, x_p \in V(G_{u,v}^+)$ where $x_1 \in S_u$, $x_p \in S_v$, $p < |E(P')|/2 - |S_v| - |S_u|$, and every two consecutive vertices lie on a common face. Consider a shortest such sequence, which visits each face of $G_{u,v}^+$ at most once. In particular, $x_1$ and $x_p$ are the only vertices of $S_u \cup S_v$.