Flow graph that requires pushing back flow in Ford Fulkerson
I think it is possible to reason backwards from an optimal solution$^1$ to the order in which the paths have to be augmented to not have any pushing-back:

In the solution, for any path from the source to the sink, where the flow is larger than zero, there is at least one edge, that constrains that flow, i.e: its flow is at capacity.
If it were not, there would have been an augmenting path along that path, i.e. the solution would not be optimal.
Also, of the edges that are at capacity, there has to be one (or multiple) with the lowest possible capacity. Furthermore, this edge has to be part of some path between source and sink where each edge has a flow at least as high. 
We take such a path, call it $P_1$ and remove its 'flow capacity' from the initial graph (i.e. reduce the capacity of the edges along the path).
We repeat this with the remaining graph (resulting in $P_2, P_3, \ldots, P_n$) until there is no path between source and sink remaining, where all edges have a non-zero capacity.

Then, $P_1, P_2, \ldots, P_n$ is the order in which the paths have to be chosen s.t. there is never any pushing back necessary. 

$1$: This only holds for optimal solutions that have been found with the FF algorithm. E.g. in a graph where 5 edges with capacity 2 go into node X and only one edge leaves it with capacity 5, there would be a max flow where each of the 5 edges has a flow of 1/2 and the outgoing has a flow of 5/5. Then the 5/5 edge would be the lowest edge at capacity but there is no path from source to sink in that graph where each edge has a flow of 5 or more. This is no problem for the algorithm outlined above because such a max-flow solution would not be found with FF.