For $k>1$, a period-$k$ orbit of a map $F$, or $k$-cycle, is a set of $k$ distinct points $\{x_0,x_1,\ldots,x_{k-1}\}$, where $x_i=F^i(x_0)$.
The part I do not understand in the above definition is how to interpret/read the notation $$x_i = F^i ( x_0 ).$$
For example, the condition for two distinct fixed points, say $U_1,U_2$, to be a $2$-cycle of $F$ is $$F(F(U_1))=U_1~\text{and}~F(F(U_2))=U_2.$$
So, the $2$-cycle is $\{U_1,U_2\}$.
Is that correct?
The notation $$x_i = F^i ( x_0 ).$$
defines $x_i$ to be $i^{th}$ iterate of $x_0$ under the function $F$, the result obtained by evaluating $F$ at $x_0$, then evaluating $F$ at $F(x_0)$, then then evaluating $F$ at $F(F(x_0))$, then continuing this process until the function has been evaluated $i$ times. If there exists any $i$ so that $x_i = x_0$, then $F$ has an $i$-cycle at $x_0$.
The notation is an abuse of the ordinary notation of exponentiation of numbers, where $a^nb$ means multiply $b$ by $a$, $n$ times. It helps to think of the case where $F$ is a linear map. Then $F$ has a matrix representation, and in this case the two notations (multiplication and composition) are identical.
For example, $$x_3 = F^3 ( x_0 ) = F(F (F( x_0 )))$$
The condition for $F$ to have a 2-cycle at any point $x$ is $$F(F(x))=x.$$ $x$ need not be a fixed point. If $F$ is not a fixed point, this requires that there exists an $a \neq x$ in the domain of $F$ so that $F(x) = a$ and $F(a) = x.$ Then $F(F(x)) = F(a) = x,$ and so $F$ has a 2-cycle at $x$ given by $\{x,a\}$. Notice that $F$ also has a 2-cycle at $a$ with the same orbit.
If $x$ is any fixed point of $F$, then $F(x) = x$, so the condition above requires that $a = x$. The orbit of the (trivial) 2-cycle is thus given by $\{x\}.$