 Solution

Keep the paper's $R_1,R_2>0$ as magnitudes for the two convex faces. With the standard signed-curvature convention measured to the right, the first surface has signed radius $+R_1$ and the second $-R_2$. This explains why the requested formula contains a sum of the two reciprocal magnitudes.

Let the axial lens thickness be $\ell$, and let $d_1$ be the distance from the first vertex to the intermediate focus that its refracted rays would reach without the second surface. The first refraction at a spherical surface givesAt the second face those rays are still converging toward that intermediate focus, so its object distance is negative, $u_2=-d_2$, with $d_2=d_1-\ell$. Applying the signed spherical-surface equation again givesIn the thin lens limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:The focal length is the image distance for parallel incident light rays, $u\to \infty$, hence the lensmaker's equation isThe same surrounding refractive index $n_1$ is assumed on both sides. For $n_2>n_1$ this biconvex thin lens has positive focal length. With signed radii $R_a=R_1$, $R_b=-R_2$, the equivalent familiar expression is $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.