Modification of Schwarz-Christoffel integral
First of all, there is no real difference between the two formulas in your post: in the second, $C$ includes $\prod (-z^k)^{\alpha_k-1}$, which is a constant.
It is somewhat remarkable that the transformation between the disk and the plane preserves the general appearance of the S-C formula. (Not so for other domains, such as an infinite strip). The steps of this transformation can be found on pages 192-193 of Conformal mapping by Zeev Nehari (published by Dover, highly recommended). I paraphrase the text with occasional relabeling of indices, as I hate seeing $\nu$ used as an index... For the halfplane, Nehari has
$$f(z)=\alpha\int_0^z \prod_{k=1}^n (z-a_k)^{-\mu_k}\,dz+\beta \tag{50}$$
By the linear transformation
$$z=i\frac{1+\zeta}{1-\zeta},\quad \zeta=\frac{z-i}{z+i}$$
which maps $|\zeta|<1$ onto $\operatorname{Im}z>0$, we can also obtain from (50) a formula for the conformal map of the unit circle onto the polygon $D$. We have
$$(z-a_k)^{\mu_k} = \left[ i\frac{1+\zeta}{1-\zeta} -a_k\right]^{\mu_k}
=\left(\frac{a_k+i}{1-\zeta}\right)^{\mu_k} \left[ \zeta - \frac{a_k-i}{a_k+i} \right]^{\mu_k} \tag{a}$$
and
$$dz=\frac{2i\,d\zeta}{(1-\zeta)^2}\tag{b}$$
Let $$b_k=\frac{a_k-i}{a_k+i} $$ which is the point on the unit circle corresponding to $a_k$. Since $\sum \mu_k=2$, the denominators $(1-\zeta)$ in (a) cancel the denominator in (b). We end up with
$$f(\zeta)=\alpha_2\int_0^\zeta \prod_{k=1}^n (\zeta-b_k)^{-\mu_k}\,d\zeta+\beta_2 $$
where $\alpha_2,\beta_2$ are constants.