Negative cycle detection using Bellman-Ford and its correctness
If there are no negative cycles then you can indeed perform the relaxation until all values remain the same. 
In the $i$-th round of relaxation you are ensured to have correctly computed all finite distances towards all vertices whose hop-distance from $s$ is at most $i$ (the hop distance is the number of edges in a shortest path).

Therefore stopping after the $(n-1)$-th iteration ensures that:


All finite distances have been correctly computed
The algorithm terminates on graphs with negative cycles.


Notice that there are graphs where all $(n-1)$ iterations are needed to properly compute the distances. E.g., a path.

Regarding 2), your question is not completely clear to me. In the proof there is no particular edge to visualize. The proof just shows that if it impossible for these two conditions to be simultaneously true:


There is some  negative weight cycle $C$.
No relaxation happens, i.e., $v.d \le u.d + w(u,v)$  for all vertices $v$ of the cycle. This is used in the second centered inequality.