Winding number of $f$ is equal to $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$ 
Let $f: S^1 \to \mathbb{R}^2 - \{0\}$ be a smooth map. Define the winding number of $f$ about 0 and prove that it equals $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

I have no clue, except for the winding number of $f$
around 0 is just the degree of $u$, where
$$u(x) = \frac{f(x)}{|f(x)|}.$$

Degree. $Y$ is connected and has the same dimension as $X$. We define the degree of an arbitrary smooth map $f: X \to Y$ to be the intersection number of $f$ with any point, $\deg(f) = I(f,\{y\}).$
Intersection number. If $f: X \to Z$ is transversal to $Z$, $f, Z$ are appropriate for intersection theory, then $f^{-1}(Z)$ is a finite number of points, each with an orientation number $\pm 1$ provided by the preimage orientation.

Thank you~

I am sorry for having no time to write a complete answer. But I suggest you to read the definition of degree in more detail (what does it tell you about the behavior of $u$?). The wikipedia article on winding number and degree may be a good start. 
