Let p be a prime element, with P the associated prime ideal, in a pid R. (Sorry for the overuse of the letter p.) Let S be the integral ring in the field extension E/F. Let Q be a prime ideal in S lying over P. (Technically, we started with Q, then contracted to P.) Note that Q contains the element p, and this will be one of our two generators.
Localize about P, whence QP becomes one of finitely many prime ideals, and SP is a pid. Let w generate QP. There is no harm in clearing the denominator, so that w lies in S. This will become the second generator. If it also generates p, then we don't need p after all.
Since Q is the pullback of QP, the elements of Q are the numerators of the elements of QP, which are the numerators for any given denominator, such as 1. This is the contraction of the extension of the principal ideal w*S. Write a/1 = x/b for some x in w*S. Thus Q consists of w*S, and any quotients thereof by elements not divisible by p.
Let w generate g*v, where g is in S and v is in R-P. Thus g lies in Q. If w does not generate g directly, there is more work to do.
Since R is dedekind and P is maximal, choose u in R such that u*v = 1 mod p. That is, uv = 1+px. Multiply u by vg, and then subtract pxg. The result is g, hence w and p span g, and w and p generate Q.