For $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z})$ the modular discriminant $$\Delta(z)=(2\pi)^{12}\eta(z)^{24}\qquad(1)$$ holds $$\Delta\left(\dfrac{az+b}{cz+d}\right)=(cz+d)^{12}\Delta(z),\qquad (2)$$ where $\eta$ is the Dedekind eta function.
I found:
Extracting 24th roots in $(2)$ and using $(1)$ we find $$\eta\left(\dfrac{az+b}{cz+d}\right)=e^{\pi i/12}(cz+d)^{1/2}\eta(z).\qquad (3)$$
Why do we have the factor $e^{\pi i/12}$ in $(3)$?
Remember that if you take $n$-th roots of an equation, there are $n$ (complex-valued) solutions corresponding to the $n$ roots of unity. So taking $24$-th roots of unity of \[\Delta(\gamma z) = (cz + d)^{12} \Delta(z)\] for $\gamma = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$, we must have that \[\eta(\gamma z) = \epsilon(\gamma,z) (cz + d)^{1/2} \eta(z)\] for some $24$-th root of unity $\epsilon(\gamma,z)$ that possibly depends on $\gamma$ and $z$. In the wikipedia page you link, it is shown that $\epsilon(\gamma,z) = e^{bi \pi/12}$ if $c = 0$ and $d = 1$, and $\epsilon(\gamma,z)$ is something more complicated otherwise.