a=3.236 gives a superstable 2-cycle i.e. a 2-cycle with one point at the maximum of f. In the demonstration the transient behavior is eliminated by iterating 100 times before plotting.
In f 2(x) the two intersection points are at stationary points of f 2 - the slopes at the two intersection points are necessarily equal by the chain rule giving the derivative of f 2 as the product of the derivatives of f at the two orbit points x1, x2.
Notice that the region near x=0.5, if inverted, resembles the quadratic map, and the behavior of the orbit starting from x=0.1 is reminiscent of the behavior in f(x) for a=2 (demonstration 1).
Starting from a slightly different initial condition (x=0.29) leads to an orbit that converges to the other fixed point:
Although this looks like a quite different orbit in f 2 this is just because we are "strobing" the orbit of f, looking at the value every other step. In fact this initial condition gives almost the same orbit in f, since f(0.1)=0.29.
The important point is that the behavior for x>0.7 i.e. on the greater side of the unstable fixed point is just a reflection of the behavior for x<0.7 (at least after a few transient steps to bring the orbit into this attracting region). This means we can focus attention on the region around x=0.5 (in fact at this value of a on the range |x-0.5|<0.2).