To identify the $T^{2}$-action we examine the Hamiltonian vector field action. Recall that $\{\frac{\partial}{\partial\theta_{i}}\}_{i=1,2}$ is dual to the connection $\{\vartheta_{i}\}_{i=1,2}$. We compute

in particular

from which the first circle action is clear. Likewise with the second circle action.

The holomorphic (3,0)-form $\Omega$ is uniquely determined by the condition that $\Omega(\frac{\partial}{\partial\theta_{1}},\frac{\partial}{\partial\theta_{2}},\cdot)=d\eta$. But

where in the last step we used the functional equation $z_{0}z_{1}z_{2}=\eta$. This shows $\Omega=-dz_{0}\wedge dz_{1}\wedge dz_{2}$. In particular the map $M\setminus\{0\}\to\mathbb{C}^{3}\setminus\{0\}$ is locally invertible.

To show the map is a homeomorphism, we notice that it is compatible with the fibration structure $M\to\mathbb{C}_{\eta}$ and $\mathbb{C}^{3}\to\mathbb{C}_{\eta}$, so it suffices to show the fibres are identified, which follows from looking at the complexification of the $T^{2}$-action into a $(\mathbb{C}^{*})^{2}$-action.

Combining the above proves the biholomorphism claim. ∎