For the "prove $f$ is a bijection" part (i.e. #1): We first show $f$ is injective. So, suppose $f(a) = f(b)$, where $a, b \in D$. Now, for $a$ and $b$ there exist positive integers $n$ and $m$ such that $f^n(a) = a$ and $f^m(b) = b$ (where $f^n$ is the standard notation for iteration: $f^n = f \circ f \circ f \circ \cdots \circ f\ (n\ \mathrm{times})$, and $f^0 = \mathrm{id}$, the identity function.). These equations give $f^{n-1}(f(a)) = a$ and $f^{m-1}(f(b)) = b$. Now, since $f(a) = f(b)$, we have $f^{m-1}(f(a)) = b$. That is, $f^m(a) = b$. Now also, $f^{n-1}(f(b)) = a$ thus $f^n(b) = a$. What $f^m(a) = b$ shows is that the cycle containing $a$ also contains $b$, thus the two cycles are not disjoint, and are in fact the same cycle. Thus, $f^n(b) = b$ as well (note that a period of one point in a cycle is a period of every point), and so $a = b$.
Now to show $f$ is surjective. Suppose $a$ is an element of $D$. Then, we want to find a $b$ for which $f(b) = a$. Since $f^n(a) = a$ for some $n$, we have $f(f^{n-1}(a)) = a$, and so $b = f^{n-1}(a)$ will work.
So, $f$ is both injective and surjective. Thus, $f$ is bijective.
For the "converse" part (i.e. #2): Suppose $f$ is bijective. Now, consider $f^n(a)$ for some $a \in D$ and positive integers $n$. Now, there are only finitely many possibilities for $f^n(a)$, so there must be some n for which $f^n(a) = f^m(a)$ for some m <= n. Then, since $f$ is bijective, it is invertible, and we have the existence of the negative iterate $f^{-m} = (f^{-1})^m$, which is the inverse of $f^m$, and we can apply that to both sides to get $f^{n-m}(a) = a$.
Q. E. D.