Prove that regular languages are closed under the cycle operator
The idea is to decide nondeterministically at the beginning how much the word is cycled, and have a copy of the automaton for every case. In terms of the automaton, that means that we guess in which state $D$ would have been after consuming a word's prefix (which is a suffix of our input), and start in that state.

Now the construction. For every state $q \in Q$, separate $D$ into two parts $A_1$ and $A_2$. $A_1$ contains the states from which $q$ is reachable and $A_2$ the states that are reachable from $q$:


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Note that any given node may be contained in both $A_1$ and $A_2$. Therefore,  the number of states can double if we make this step explicit.

Now we rewire this automaton so it accepts all words for which $q$ marks the "cycle point":


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We get $|Q|$ automata of this form; create a new initial state which has $\varepsilon$-transitions to all their starting states. The resulting automaton accepts $\operatorname{cycle}(L)$. Altogether, we get at most $|Q|\cdot(2|Q|+1) + 1$ states, only $|Q|$ more states than the reference claims are possible.

You can achieve $2|Q|^2 + 1$ states by modifying the component automata a little bit; eliminate all $q_0$ by replacing the incoming $\varepsilon$-transitions with copies of its outgoing transitions. That is, for every pair of transitions $(q_1,\varepsilon,q_0), (q_0,a,q_2)$, introduce a transition $(q_1,a,q_2)$.

Rigorous construction and correctness proof remain as exercise.