Ford-Fulkerson algorithm clarification
Prior to step (c) there are 7 units flowing from $v_1$ to $v_2$. The augmenting path pushes 8 units of flow from $v_2$ to $v_1$: 7 of these cancel the existing flow from $v_1$ to $v_2$, and 1 unit now flows from $v_2$ to $v_1$.

More generally, the residual network shows you how much flow you can push in each direction. The edge $v_2 \to v_1$ has a capacity of 11, so you can push up to 11 (=7+4) units in this direction.