Why is augmenting path calculation O(m)
As given in your lecture notes, and an augmenting path is an s-t path in the residual graph such that all edges on the path have a positive residual capacity.
There could be many augmenting paths in the graph. But you just need to find one such path. Therefore, you can apply DFS or BFS which takes $O(m)$ time to find one such path from $s$ to $t$.