I think this is very, very simple and obvious, but I can not understand the problem. I guess it is just a typo.
In this context, $F$ is a real quadratic field, and $x\mapsto x'$ stands for the conjugate of $x$ in $F$. At the beginning of the second chapter of the book "The 1−2−3 of Modular Forms", a lemma is stated about the Imaginary part of $z=(z_1, z_2)$ (Note that $a', b', c', d'$ are conjugates of $a, b, c, d$):
The lemma says $\mathfrak{I}(\gamma(z))= \Big(\dfrac{\mathfrak{I}(z_1)}{|N(cz+d)|^2}, \dfrac{\mathfrak{I}(z_2)}{|N(cz+d)|^2}\Big)$, but my calculations say something else:
$$\mathfrak{I}(\gamma(z)) =\Big(\mathfrak{I}(\gamma(z_1)), \mathfrak{I}(\gamma'(z_2))\Big) =\Big(\frac{\mathfrak{I}(z_1)}{|cz_1+d|^2}, \frac{\mathfrak{I}(z_2)}{|c'z_2+d'|^2}\Big),$$
for example for the first coordinate, we multiply the numerator and the denominator by $c\bar{z_1}+d=\overline{cz_1+d}$. Then
$$\mathfrak{I}(\gamma(z_1))=\frac{az_1+b}{cz_1+d}=\frac{az_1+b}{cz_1+d} \times \frac{c\bar{z_1}+d}{\overline{cz_1+d}} =\frac{(ac|z_1|^2+bd)+(adz_1+bc\bar{z_1})}{|cz_1+d|^2}=\frac{(ac|z_1|^2+bd)+(adx_1+bcx_1)+i(ad-bc)y_1)}{|cz_1+d|^2},$$
and note that the first two parentheses in the numerator of the last fraction are real, also $ad-bc=1$, and $\mathfrak{I}(z_1)=y_1$, and we are done.
I can not understand where is the problem? Am I misunderstanding something?