There are two parts of your proof that I would consider incorrect, leaving out important details (which I'm sure you were aware of, but didn't write down):
In the $P \implies Q$ direction, Case 2, you only show that $\deg v_0 \ge 2$, which doesn't mean that it's even: $17 \ge 2$ and $17$ is odd. What you really want to argue here is that we use edges out of $v_0=v_k$ once at the very beginning, once at the very end, and twice every time it appears in the middle of the walk (this part is identical to Case 1), for a total of $2n+2$ edges, which is even.
In the $Q \implies P$ direction, when you write
we branch into $w_j$, walk until we reach $u_i$ again and continue the walk
this will not generally produce an Eulerian tour at the end. The closed walk $w_j$ might visit $u_i$ many times, and if you continue along $c$ the first time you return to $u_i$, you might miss out on many edges in $G_j'$. Instead, what you should do in this step is walk the entirety of the closed walk $w_j$, starting at $u_i$ and therefore ending at $u_i$, and then continue along $c$ to $u_{i+1}$.
Aside from that, I just have a couple of comments on clarity.
First, when you begin your induction
Let $P(n)$ be the proposition that if all $v\in V(G)$ have even degree, then a connected graph $G$ has an Euler circuit. We induct on $n=|E(G)|$.
you're not really using the right form for $P(n)$: you haven't made it a statement that depends on $n$. What you really want to define $P(n)$ to be is the proposition that "If $G$ is a connected graph with $n$ edges and all $v \in V(G)$ have even degree, then $G$ has an Euler circuit." (This also avoids the problem that, within the statement, you talk about $V(G)$ before you've said what $G$ is.)
Second, I'm not sure $P(3)$ is the right base case. I'd actually go with $P(0)$ (a graph with one vertex and no edges is connected, all vertices have even degree, and it has a trivial closed walk) because in your induction step, when you delete the cycle $c$, you may well end up with plenty of components that $P(0)$ should apply to. (And if we have $P(0)$, we can use it to prove $P(3)$: remove the cycle of length 3, and you have three components with $0$ edges.)
Third, a minor nitpick: when you're combining all the cycles together, you should only take the closed walk of $G_j'$ the first time you reach a vertex $u_i \in c$ that is a part of $G_j'$. (This is obvious if you know what you're doing, but it needs to be said if you want to describe a correct algorithm.)
But overall, well done! This is a complicated proof to write, with lots of fiddly details (which I didn't fully appreciate myself before I had to teach a recitation on it last semester), and you did a good job hitting all the important points. Don't let the length of my comments make you think otherwise.
Note that a graph is Eulerian if and only if it is connected and the degree of every vertex is even.
We can have many bipartite graphs with these properties.
For example $$k_{2,2}$$ is Eulerian.
Your confusion with the edges is that , you think we have to trace every potential edge in the graph as well as every edge which is actually an edge.
Note that you only have to trace the edges that exist not the potential edges which could be made by joining vertices which are not connected at this point of time.
Best Answer
The definition of extendable says that every trail starting at $v$ can be extended to an Eulerian circuit of $G$. But the Eulerian circuit we find in $G - E(C)$ is a trail starting at $v$ which cannot be extended any further, contra the definition.
In more detail: removing the cycle $C$ only changes degrees by $2$, therefore, by the familiar necessary and sufficient condition for the existence of an Euler circuit, as $G$ was Eulerian, and so had all vertices of even degree, $G - E(C)$ has all vertices of even degree and so each connected component of $G - E(C)$ is Eulerian. This provides an Euler circuit $R$ in the component of $v$ in $G - E(C)$ - that is, a closed trail which passes through every edge. In particular, we may choose this trail to start (and so end) at $v$; and in particular it passes through every edge incident to $v$ (note that this includes every edge incident to $v$ in the original graph $G$, as the cycle $C$ that we removed lies in $G - v$).
Then, in $G$, we may start at $v$ and follow $R$ round back to $v$ again. We have used up every edge incident to $v$, but we have have not visited any edge of $C$. So the trail $R$ cannot be extended to an Euler circuit of $G$.