Given that iterating $f$ $m$ times are written like \begin{align*} f^m(x)=(f\circ f\circ \ldots \circ f)(x) \end{align*} and $n$th derivative is written like \begin{align*} f^{(n)}(x)=f\underbrace{''^{\ldots}{'}}_{n~'s}(x) \end{align*}
Is there a way to represent the $n$th derivative of $f^m(x)$ in a single expression and vice versa?
There exists very easy way.
We can denote n-th derivative of $f$ as $\frac {d^n}{dx^n}f$.
So...
$$ \frac{d^n}{dx^n}f^m(x) $$
can be the answer.