In the following max flow problem, the algorithm could choose S-A-D-T path first. In that case, the algorithm will not see any more augmenting paths so it will produce 4 as the answer to the maximum flow. However, if the algorithm would select any other path first, then it would see that the max flow becomes five.

It seems you did not notice that Ford-Fulkerson algorithm updates the residual graph in two ways:
Therefore after you have selected the
S-A-D-Tpath with flow3, the residual graph will now contain an augmenting pathS-B-D-A-C-Twith flow2, thus reaching a total flow of5. Flow along the reverse edges just cancels the existing flow and thus only flow of2will remain fromAtoD.