Curvature function and rate of change of angle 
Let $\gamma:(a,b)\rightarrow \mathbb{R}^2$ be a smooth curve with $\| \dot{\gamma}(s)\|=1$ for all $s\in (a,b)$. Fix $s_0\in (a,b)$ and let the unit vector $\dot{\gamma}(s_0)$ be represented by $(\cos \phi_0,\sin\phi_0)$.  Then there is smooth function $\phi$ with $\phi(s_0)=\phi_0$  such that 
  $$\dot{\gamma}(s)=(\cos\phi(s),\sin\phi(s))$$
  for all $s\in (a,b)$. 

The proof goes as follows: let 
$$\dot{\gamma}(s)=(f(s),g(s))$$
so that $f(s)^2+g(s)^2=1$ for all $s$. Define 
$$\phi(s)=\phi_0 + \int_{s_0}^s (f\dot{g}-g\dot{f})du$$
It is then shown that this is required $\phi$ in the theorem.
Q. I didn't get intuition for choice (definition) of $\phi$. How do we justify the choice of $\phi$ above? 
Reference: Elementary differential geometry by Pressley, Proposition 2.2.1 (New edition)

Using the explicitly defined angular function $\phi$, the curvature function is given by 
$$\kappa_s =\frac{d\phi}{ds}.$$

Intuition is that you want to get $\phi$ by integration of $\kappa=\frac{d\phi}{ds}$.
You get $$\phi(s)=\phi_0 + \int_{s_0}^s \kappa(u) du$$
by fundamental theorem of calculus. From serret formulae you can derive $$\kappa=\langle \ddot  \gamma,J \dot \gamma \rangle$$ 
where $\dot\gamma$ interpreted as tangential vector field $T$ and $J$ a rotation anticlockwise by $1/2 \pi$.
This will yield you 
$$\phi(s)=\phi_0 + \int_{s_0}^s \langle \dot T(u),JT(u) \rangle du=\phi_0 + \int_{s_0}^s (f\dot{g}-g\dot{f}) du$$ as desired.
