Let $f:[0,1]\to [0,1]$ be continuous with a period three orbit. We know with Sharkovskii`s theorem that f owns period orbits of any natural number.
If we take a look to the bifurcation-diagram of the logistic map, we can find a period three orbit for e.g. $r=3.83$, see bifurcation-diagram and period three orbit of the bifurcation-diagram.
To give an example: Let $f:[0,1]\to [0,1]$, $x\to 3.83x(1-x)$ and $g(x):=f(f(f(x)))$ the triple composition of f. The bifurcation-diagram of the logistic map shows a period three orbit at $r=3.83$, so g has three fixed points of period three and f has three accumulation points. With Sharkovskii`s theorem I'd expect orbits of all periods at this given $r=3.83$. The bifurcation-diagram only shows three stable fixed-points. Wikipedia says: " In fact, there must be cycles of all periods there, but they are not stable and therefore not visible on the computer-generated picture." But if I plot the graph of g, we only find a finite number of fixed points (both stable and unstable) but not infinite, see graph of $g$.
So how can I explain, that this doesn't contradicts with Sharkovskii`s theorem? How is it possible, that g has a infinite number of fixed points if it only has a finite number of intersections with the bisecting line, wich are fixed points?
The periodic orbits of higher period (say $n>3$) correspond to fixpoints of the $n$-fold iterated map $f^{(n)} = f \circ f \circ \dots \circ f$, not of the $3$-fold iterated map $g = f^{(3)} = f \circ f \circ f$. So you should look for them in the graphs of $f^{(4)}$, $f^{(5)}$, $f^{(6)}$, etc. – not in the graph for $g$.
(As you say, obviously $g$ doesn't have infinitely many fixpoints. But nobody is claiming that it should! "There must be cycles of all periods" means something completely different.)