That is simply not true.
Indeed for $r=4$, the graph of $f^n=\underbrace{f\circ f\circ \cdots\circ f}_{n\text{ times}}$ is composed of $2^{n-1}$ arches with $y$-coordinates $0\rightarrow 1\rightarrow 0$.
Therefore the graph of $y=f^n$ and $y=x$ has $2^n$ intersections and so $2^n$ period-$n$ points.
In fact the periodic points of that map are dense in $[0,1]$ for $r=4$.