2. Given the function
$$
\begin{array}{l}
f(x)=f_{1}(x) \\
\quad=\left\{\begin{array}{ll}
16(x-0.25)^{2}, & 0 \leqslant x<0.5 ; \\
16(x-0.75)^{2}, & 0.5 \leqslant x \leqslant 1 .
\end{array}\right.
\end{array}
$$

When $n \geqslant 2$,
$$
f_{n}(x)=f\left(f_{n-1}(x)\right)(x \in[0,1]) \text {. }
$$

Then the number of real solutions to the equation $f_{2012}(x)=\frac{1}{3} x$
$2.4^{2012}$.
It is easy to know that the domain and range of the function $f(x)$ are both the interval $[0,1]$. From the graph of $f(x)$, we know that the domain of $f(x)$ can be divided into 4 monotonic intervals, meaning that when $x$ runs through $[0,1]$ once, $f(x)$ runs through $[0,1]$ four times.
Further induction leads to:
The domain of $f_{2012}(x)$ can be divided into $4^{2012}$ monotonic intervals.

Therefore, the graph of $f_{2012}(x)$ intersects the line $y=\frac{1}{3} x$ at exactly $4^{2012}$ points, meaning the number of real solutions to the equation $f_{2012}(x)=\frac{1}{3} x$ is $4^{2012}$.