Bounding the derivative of a holomorphic function at the origin.
So $g : \mathbb{D} \to \mathbb{D}$ satisfies $g(1/2)=0$. Then using your Möbius transformation you can define $h : \mathbb{D} \to \mathbb{D}$ by $h(z) = g((2z+1)/(2+z))$ ($\theta=0$ and $z_0=-1/2$) so that $h(0)=g(1/2)=0$. Note that there is some $w_0$ such that $h(w_0)=g(0)=f'(0)$. So you can get what you want from the Schwarz lemma by solving $w_0$.
$$p(z)=\frac{f(z)}{z \frac{z-1/2}{1-z/2}}$$ is analytic on the unit disk and $|p(z)|\le 1+\epsilon$ on $|z|=1-\delta$ so $|p(z)|\le 1$ on $|z|<1$. Then note that $$p(0)=\frac{f'(0)}{-1/2}$$