No cycle containing edges $e$ and $g$ implies there is a vertex $u$ so that every path sharing one end with $e$ and another with $g$ contains $u$ There is a proof in my textbook for the following claim, which doesn't make a whole lot of sense to me.  My annotations are in bold.  Could someone perhaps elaborate on what's going on?
Claim.  If there does not exist a cycle containing edges $e$ and $g$ then there exists a vertex $u \in V (G)$ such that every path in $G$ sharing one end with e and another with $g$ contains $u$.
Proof: The claim trivially holds if $e$ or $g$ is a loop OK, so we assume that neither is. Let $P$ with vertex set $v_1, v_2, . . . , v_k,$ in order, be a path with $e$ joining $v_1$ to $v_2$ and $g$ joining $v_{k−1}$ and $v_k$. Let $f_i \in E(P_i)$ be the edge with ends $v_i$ and $v_{i+1}$. Let $j$ be chosen minimum so that no cycle in $G$ contains $e$ and $f_j$ We can do this because we know that at least the edge $g$ will not create a cycle by assumption, right?. We will show that $u = v_j$ satisfies the claim.
Suppose not. Let $C$ be a cycle containing $e$ and $f_{j−1}$ What if $f_j = e$?  then what cycle? a single vertex? and let $P′$ be a
path from an end of $e$ to an end of $f$ Not sure what the book meant by $f$ here, any guesses? avoiding $u$. Choose a subpath $Q$ of $P′$ with one end in $V (C)$ and another in ${v_{j+1}, v_{j+2}, . . . , v_k}$ as short as possible. Then $C \cup Q \cup P$ contains a cycle containing both $e$ and $f_j$, a contradiction. (The last statement requires some case checking.) This ending seems a bit abrupt and non-obvious to me
Thanks for the help
 A: The answer to your first question is yes: we know that even if no smaller $j$ works, $j=k-1$ does. 
Your next question identifies a genuine sloppiness in the argument. If $e$ is not part of any cycle, then $j=1$, and the argument has to be made differently. Suppose that $e$ is not part of any cycle. If there were a path from $v_1$ to either end of $g$ that did not use $e$, it together with $P$ would contain a cycle that included $e$, contrary to hypothesis. Thus, any path sharing one end with $e$ and the other with $g$ must contain $v_2$; i.e., we can take $u=v_2$.
With that special case out of the way we can assume that $j\ge 2$, and we’ve set $u=v_j$. We’re supposing that we have a cycle $C$ that contains $e$ and $f_{j-1}$ (which will also be $e$ if $j=2$). The book’s $f$ must be $g$: in order to get a contradiction we’re assuming that there is a path from one end of $e$ to one end of $g$ that avoids the vertex $u=v_j$. We know that $e$ is in the cycle $C$, so $P'$ has at least one vertex in $V(C)$. We also know that $P'$ includes an end of $g$ (once the typo is fixed), so it includes at least one of the vertices $v_{j+1},\ldots,v_k$, and we can choose $Q$ as described.
The ending is a bit abrupt, presumably because the author wants the reader to have the practice of filling in the necessary details. I’ll add an explanation, but I think that I’m going to have to make some sketches to accompany it, so it may be a little while.
Added: Here’s the basic sketch:

The horizontal black line represents $P$, the blue and red together are $P'$, the blue is $Q$, and the letter $C$ is sitting inside the cycle $C$. In this sketch if you start at $v_2$, the righthand end of $e$, follow $Q$ to the point at which it meets $P$, and head back along $P$ to $u$, you’ll use the edge $f_j$ to get to $u$. To complete the cycle so that it contains $e$, take the lower part of $C$ to $v_1$ and then close the cycle by taking $e$ to $v_2$.
This sketch goes with the case in which $P'$ goes from $v_2$ to an end of $g$. If $P'$ had gone from $v_1$ instead, you’d have had to use the top part of $C$ in the picture. In other words, in one case you want the part of $C$ that runs from $u$ to $v_1$ without using $e$ or $f_{j-1}$, and in the other case you want the part of $C$ that runs from $v_{j-1}$ to $v_2$ without using $e$ or $f_{j-1}$. 
And of course it’s possible that the cycle $C$ minus the edges $e$ and $f_{j-1}$ actually has pieces that run from $u$ to $v_2$ and from $v_{j-1}$ to $v_1$ instead, hooking up the ends of $e$ and $v_{j-1}$ the other way around. That gives you another two cases, depending on the piece of $C-\{e,f_{j-1}\}$ to which $Q$ is attached. (In this sketch it’s the upper piece, running between $v_2$ and $v_{j-1}$.)
That’s certainly not a complete, detailed treatment of the cases, but it should at least give you a good start.
