The action on $\mathbb{C}^3$ given by $(e^{i\theta_1}, e^{i\theta_2}) \cdot (z_0, z_1, z_2) = (z_0, e^{i\theta_1}z_1, e^{i\theta_2}z_2)$ has for its moment map $\mu(z_0, z_1, z_2) = \frac{1}{2} (|z_1|^2, |z_2|^2)$ (up to sign which depends on convention).
This action preserves the unit sphere $\lbrack |z_0|^2 + |z_1|^2 + |z_2|^2 = 1 \rbrack$ and commutes with the Hopf fibration $\pi : S^5 \to \mathbb{C}P^2 : (z_0, z_1, z_2) \mapsto [z_0 : z_1 : z_2]$, so it induces an action on $\mathbb{C}P^2$ which is precisely the one you are considering.
You can restrict the above map $\mu$ to $S^5$ and observe that it coincides (on $S^5$) with the map $\mu'(z_0, z_1, z_2) = \frac{1}{2} \, \left( \frac{|z_1|^2}{|z_0|^2 + |z_1|^2 + |z_2|^2} \, , \, \frac{|z_1|^2}{|z_0|^2 + |z_1|^2 + |z_2|^2} \right)$. Observe that $\mu'$ is constant on the fibers of the Hopf fibration ; In fact, it is invariant under the action of $\mathbb{C}^{\times}$ on $\mathbb{C}^3$ which yields $\mathbb{C}P^2$ as a quotient. Therefore, $\mu'$ determines a well-defined map $M$ on $\mathbb{C}P^2$.
In order to prove that $M$ is the moment map for the induced action on the projective space, the best is to lift everything on the sphere. Observe that $\mu' = \pi^{\ast}M$, that the 'symplectic' form on $S^5$ (induced by its inclusion in $\mathbb{C}^3$) is the pullback by $\pi$ of the Fubini-Study form on $\mathbb{C}P^2$, that $\mu'$ is the 'moment map' for the 'Hamiltonian' action of $\mathbb{T}^2$ on $S^5$ and that the Hamiltonian vector field $\xi^{\sharp}$ on $\mathbb{C}P^2$ associated to any element $\xi = (\xi_1, \xi_2) \in (Lie(\mathbb{T}^2))^{\ast}$ is $\pi$-related to the 'Hamiltonian' vector field $\xi^{\sharp \, '}$ on $S^5$ associated to the same $\xi$. Therefore, the fact that $\mu'$ is a 'moment map' for the 'Hamiltonian' torus action on the sphere (i.e. $d\mu'(\xi) = \iota_{\xi^{\sharp \, '}} \, \omega$) implies that $M$ is the moment map of the Hamiltonian torus action on $\mathbb{C}P^2$ (i.e. $dM(\xi) = \iota_{\xi^{\sharp}} \, \omega_{FS}$).