Begin looking at the graph at a point where y' = 0 -- i.e. the curve's tangent line has a slope of 0. Here, there is a peak number of predators and n/u adult prey. At this point in time, the predator population begins to decrease, because the prey population is no longer sufficient to support that peak number. [This happens in the "second quadrant" of the system, if you place the origin of these imaginary axes at the equilibrium.]
When the system reaches the point on the left side where a' = 0 -- i.e. the curve's tangent line is vertical -- the predator population has decreased so much that the effect of predation on the adult prey population is minimal. During the time when the population of the predators is so small, the children who had been maturing while the adults were being eaten begin to "grow up" and become officially adults. [This happens in the "third quadrant"]
At the second point where y' = 0 -- the lowest point of this cycle -- the children have sufficiently repopulated the prey population that the y population can begin to increase again (y' goes from being negative to being positive). [This happens in the "fourth quadrant"]
At the point on the right side where a' = 0, the predator population is again large enough to be a drain on the prey population (a' goes from positive to negative) and the prey population begins to decrease. [This is the "first quadrant"]
When the system reaches a horizontal tangent line again, the predators have eaten so many of the prey that there is no longer enough food to support them all, and they begin to die off. This begins the "second quadrant" phase again, though on an orbit closer to the equilibrium than the last cycle was.
One can give a similar explanation for each of the three 2-D graphs.
Note that when a' = 0, y is = to the y-coordinate of the equilibrium, and when y' = 0, a' is = the a-coordinate of the equilibrium.