Is this graph $G$ 2-connected and non-hamiltonian? Does Fleischner's theorem apply to $G^2$? More specifically, does the stronger statement apply (Georgakopoulos 2009)?
"If $G=(V,E)$ is a 2-connected ﬁnite graph and $x \in V(G)$, then $G^2$
has a Hamilton cycle whose edges at $x$ lie in $E(G)$."
I am not sure I understand that statement fully. In the graph $G$ below the line $ag$ (or some other grey line) does not lie in $E(G)$, and I assume the existence of Hamilton cycle requires that line in G^2. 
I either proved Georgakopoulos wrong (yeah right) or misunderstood something badly. I got wrong the statement, hamiltonicity, definition of $G^2$, or something else.
The graph in question 
$G$ (black) and $G^2$ (black and grey):

 A: Ok as far as I see you are confused about this part of the paper.
Theorem 1. If $G$ is a 2-connected finite graph and $x \in V(G)$, then $G^2$ has a Hamilton cycle whose edges at $x$ lie in $E(G)$.
What this means is the following. Let $x \in V(G)$ be a vertex incident with the edges $\{e_1,e_2,\ldots,e_k\}$ in $G.$ Then there exist a Hamiltonian cycle $C$ in $G^2$ such that the edges incident with $x$ in $C$ are in $\{e_1,e_2,\ldots,e_k\}.$
Hopefully this clears your confusion?
A: The question has actually 3 parts, and here is answer to the two first ones.
The graph $G$ in question is 2-connected, because no two vertices can be separated by removing one vertex.
The graph $G$ is not hamiltonian. According to this paper (page 7 theorem 2), order of $S$ must be greater than or equal to the number of components of the graph $G-S$ for $G$ to be hamiltonian. 
You can think of the proof inductively: when you remove a vertex from a circle, it either keeps the number of components same or adds precisely one component to the graph $G-S$. When you remove one more vertex, number of components increase only if the removed vertex is not adjacent to any previously removed vertex.
In graph  $G$ when $S={f,b}$, number of components of the graph $G-S$ is 3, which is greater than $|S|=2$.
