Example 2 Given $f(x)=\{\begin{array}{ll}x+\frac{1}{2} & 0 \leqslant x \leqslant \frac{1}{2}, \\ 2(1-x) & \frac{1}{2}<x \leqslant 1,\end{array}$ define $f_{n}(x)=\underbrace{f(f(\cdots f}_{n \uparrow 1}(x) \cdots)), n \in \mathbf{N}^{*}$. Find the value of $f_{2006}\left(\frac{2}{15}\right)$.
Analysis From the form given in the problem, the periodicity seems obscure at first glance. However, upon closer inspection, if it were not a periodic function, how could we solve it! After calculating $f_{1}\left(\frac{2}{15}\right), f_{2}\left(\frac{2}{15}\right), f_{3}\left(\frac{2}{15}\right)$, one might lose confidence again. In fact, as long as you hold on, the solution will become clear.
Solution From the given function expression, we can obtain:
$$
\begin{array}{l}
f_{1}\left(\frac{2}{15}\right)=\frac{2}{15}+\frac{1}{2}=\frac{19}{30}, \\
f_{2}\left(\frac{2}{15}\right)=f\left(\frac{19}{30}\right)=2\left(1-\frac{19}{30}\right)=\frac{11}{15}, \\
f_{3}\left(\frac{2}{15}\right)=f\left(\frac{11}{15}\right)=2\left(1-\frac{11}{15}\right)=\frac{8}{15}, \\
f_{1}\left(\frac{2}{15}\right)=f\left(\frac{8}{15}\right)=2\left(1-\frac{8}{15}\right)=\frac{14}{15}, \\
f_{3}\left(\frac{2}{15}\right)=f\left(\frac{14}{15}\right)=2\left(1-\frac{14}{15}\right)=\frac{2}{15},
\end{array}
$$

Therefore, $f_{n}\left(\frac{2}{15}\right)$ changes with a period of 5, i.e.,
$$
f_{\text {i.k }}\left(\frac{2}{15}\right)=f_{r}\left(\frac{2}{15}\right), \text { since } 2006=5 \times 401+1 \text {, }
$$

Thus,
$$
f_{2 \text { momi }}\left(\frac{2}{15}\right)=f_{1}\left(\frac{2}{15}\right)=\frac{19}{30} \text {. }
$$

This solution process illustrates that in general, if a function or a sequence is given in an $n$-fold form, one should first consider its periodicity. Viewing sequences through the lens of functions is not uncommon, but representing the recursive relationship of a sequence using piecewise functions is relatively rare. The essence is a "sequence generator" program.
In this problem, $f(x)$ has a 5-periodic orbit $\left\{\frac{2}{15}, \frac{19}{30} \cdot \frac{11}{15} \cdot \frac{8}{15}, \frac{14}{15}\right\}$.