My intuition says it would simply be the number of edges leaving s.
I'm assuming it's a valid flow network so sum of capacities leaving s is the same as the sum of capacities entering t, so a max flow exists.
This variation simplifies the running time because for any augmenting path, we can add the same amount of positive flow which is equal to any original edge capacity. So no edge leaving s would get used twice and the runtime is simply how many edges leave s.
I think that makes sense, but maybe I'm missing something...