I know the composition of Riemann integrable functions is not necessarily Riemann integrable. But I am not finding any argument how to conclude this for self composition, or how to find a counterexample. $f$ is Riemann integrable on $[a,b]$ means that the discontinuity set of $f$ is of measure zero. But the discontinuity of the composition is a larger set, so we cannot conclude anything.
Similarly for the self composition of measurable functions: is it measurable or not? I am unable to find a counterexample.
Suppose you have Riemann integrable functions $g,h : [0,1] \to [0,1]$ such that $h \circ g$ is not Riemann integrable. Define $f : [0,3] \to \mathbb{R}$ by $$f(x) = \begin{cases} g(x) + 2, & 0 \le x \le 1 \\ 123.456 & 1 < x < 2 \\ h(x-2), & 2 \le x \le 3. \end{cases}$$ Then $f \circ f = h \circ g$ on $[0,1]$. I leave it to you to verify that $f$ is again Riemann integrable, that $f \circ f$ is not, and to make adjustments if you don't like the domain and range I chose.