Yes Ford-Fulkerson can. Always solving such problems is actually what it was designed for.
I guess the fact you are missing is that when determining the residual graph you also have to add the back edges. Thus after going 0->1->3->6 and 0->1->5->6 the residual graph actually looks like this:
1 1
+-------> 4 ------+
| |
2 | 1 1 v
0 <------ 1 <----- 3 <----- 6
| ^ |
| 1 | 1 | 1
+-------> 5 <---------------+
The next step Ford-Fulkerson is going to take is adding the route 0->5->1->4->6 thus yielding a flow of 3.