The information is enough to get the function value

We can prove that $f(x)$ is differentiable for any value of $x$ in its domain and find a differential equation for $f(x)$ by using the definition of derivative. $f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac{\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)}{h}=\lim_{h\to 0} ( \frac{1+f^2(x)}{1-f(x)f(h)}*\frac{f(h)}{h}).$

Now, using that $\lim_{h\to 0} \frac{f(h)}{h}=2$ and $\lim_{h\to 0} f(h)=\lim_{h\to 0} \frac{f(h)}{h}*h=2*0=0,$ we obtain that the function $f(x)$ has derivative at any point $x$ where it is defined (therefore, it is continuous), and $f'(x)= 2(1+f^2(x)).$ In other words, $f(x)$ is a solution of the differential equation $\frac{dy}{dx}=2(1+y^2).$ By separation of variables we get $\frac{dy}{1+y^2}=2dx.$

Integrating both sides with respect to $x$ we get $\arctan y=2x+C$ . Since $f(0)=\lim_{h\to 0} f(h)=0$ , we obtain that $y =0$ when $x=0.$ So the value of the constant $C$ must be 0. That is, $\arctan y=2x$ Solving this equation for $y,$ we obtain that $y=\tan (2x)$ , then $f(x)=\tan(2x)$ where $-\frac{\pi}{4}<x<\frac{\pi}{4}.$ So the answer is $f(\frac{\pi}{12})=\tan \frac{\pi}{6}\approx 0.58.$