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](https://physics.stackexchange.com/questions/139532/the-conjugate-representation-in-mathfraksu2) or [this](https://physics.stackexchange.com/a/534028/66086).

 1. 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).

 2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra: $\zeta^{-1} \sigma^k\zeta  =- \sigma^{k~*}$, whence $\zeta^{-1}  U\zeta = U^*$, reviewed in good texts (Li & Cheng, etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above. 

 3. 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!

 4. 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 $i(\xi_2^2+\xi_1^2)$  to  all be real.

 5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

----------------- 

 6. Extra credit. This has presumably nothing to do with your text, but now that  one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define [another 2×2 matrix](https://physics.stackexchange.com/questions/472637/embedding-of-su2-l-times-u1-y-into-su2-l-times-su2-r-in-electrowea), linear in them, whose first column is $\tilde \xi$, and the second $\xi$,
$$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ 
M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}.  $$
It is evident by above that 
$$
M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . 
$$ 
since each of its columns transforms thusly. However, imagine another SU(2) transformation, completely unrelated and oblivious to the above *U* one, acting on the *right*:
$$
M\to MV,
$$
for unitary *V*. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2), and $M^\dagger M$ transforms in the adjoint of the new SU(2), *V*.   By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.