I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.
He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).
It is a long story, and has to do with the conjugacy isomorphisms of the SU(2) Lie algebra, $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, treated in good books (Li & Cheng, etc..) and several questions on this site. Take it as a given—the "magic" above.
I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!
This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ are all real.
Not apparent with the information you provide. To be sympathetic with your bafflement, there are superior introductions to SU(2) around, even on this site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you persist?