Consider the disks of radius 1 centered at 0 and 1 in the complex plane. Their intersection forms a lens shape. I want a complex function which is a conformal map from this lens to the first quadrant.
What are the basic strategies I can use to find such a mapping?
First use a Mobius transformation sending (the intersection points of the circles) $1/2+i\sqrt{3}/2$ to $\infty$ and $1/2-i\sqrt{3}/2$ to $0$ (so that the boundary curves become straight lines, and our domain is nicer): I'll go with $$f_1(z) = \frac{z-1/2+i\sqrt{3}/2}{z-1/2-i\sqrt{3}/2}.$$ (I'm not very original.) Verify that your boundary curves map to the rays $t(-1+i\sqrt{3})$ and $t(-1-i\sqrt{3})$ (both for $0 \leq t \leq \infty$). A quick check (say, plug in $z=.75$) sees that your lens has mapped to the area to the left of these two rays. I prefer to include the real axis in my domains, so my next map is gonna be $f_2(z) = -z$ to flip stuff over.
Now our rays make an angle of $2\pi/3$ with each other; we want this to be $\pi/2$, so compose with $f_3(z) = z^{3/4}$ (taking the "principle branch", $re^{i\theta} \mapsto r^{3/4}e^{3i\theta/4}$ for $|\theta|< \pi$.) After what we've done so far, our rays are $t(1+i)$ and $t(1-i)$. To get to the first quadrant, we need to rotate: our final map will be $f_4(z) = (1+i)z$. All together, the conformal map you desire is $$f_4(f_3(f_2(f_1(z)))) = (1+i)\left(-\frac{z-1/2+i\sqrt{3}/2}{z-1/2-i\sqrt{3}/2}\right)^{3/4}.$$
The methods we used there should generalize perfectly well to any region bounded by two circular arcs.