This behavior is perhaps more easily understood by looking at the iterate of f i.e. f (2)=f(f(x)), which shows every other iterate of f. Remember that a period 2 orbit of f appears as a fixed point of f (2) (actually there are two fixed points, corresponding to the two points on the orbit of f). You can also convince yourself that a fixed point of f is also a fixed point of f (2). So set nf=1, reset a to a=2.8 and iterate: the orbit converges to the single fixed point.
Again increase a to a=3.2. There are now three fixed points (intersections of the diagonal with the curve f (2)). The orbit passes close to the "old" central fixed point, but diverges from here and approaches one of the other two fixed points (which one depends on the initial value x0). Comparing this and the previous iteration, it might be clear that the instability of the "old" fixed point is associated with the slope of the curve f (2) becoming greater than unity - a general result. It should also be apparent that as the slope passes through unity, the two new fixed points grow out from the old one. Returning to the original function f, this corresponds to the absolute value of the slope becoming greater than unity (the slope actually passes through -1) and a period two orbit growing out of the unstable fixed point.
We have seen that the period two cycle is simple (a fixed point!) in f (2). Set nf=1 and a=3.4 and iterate the orbit. The orbit passes close to the unstable "old" fixed point and converges to one of the f (2) fixed points, which are stable. Increase a to a=3.5. Now all the fixed points of f (2) are unstable and the orbit goes to a period 2 cycle of f (2) corresponding to the period 4 cycle of f.
We can vaguely see that the fixed point has gone unstable in the same way the fixed point of f went unstable. To make this even clearer expand the central region by using the rescale variable nsc=1 and iterating for a=3.4 and a=3.5.
Finally, we understood the instability of the fixed point in f by looking at f (2): similarly it is useful here to look at f (4) by setting nf=2, and iterate for a=3.4 and a=3.5. The instability of the period 2 cycle is seen to be associated with the slope of f (4) becoming greater than unity, and again two new fixed points of f (4) develop.
Clearly the process repeats, and we will find a succession of bifurcations to more and more complicated 2m cycles with m=0,1,2,...... The values of a at which these bifurcations occur get closer and closer together, so that by the time a has reached ac which is 3.569946... for the quadratic map, an infinite number of bifurcations has occurred. The complex "2 to the infinity" cycle orbit corresponds to the onset of chaos.
The sequence of bifurcations can be studied (either in the quadratic or sine maps) using the drop down boxes at the bottom of the screen: choose Quadratic or Sine, the - sign, and Instability and then the value of m in the last box gives the a for the 2m cycle to just go unstable This is best seen looking at nf=m amd nsc=m.