I have decided to rewrite my question almost entirely: Let $Y$ be a compact (without boundary) Calabi-Yau manifold, i.e., $c_1(Y)=0$ in $H^2(Y, \mathbb{R})$. Let $\omega$ be a Kähler form on $\mathbb{C}^m \times Y$ and let $\omega_P = \omega_{\mathbb{C}^m} + \omega_Y$. It can be shown (see, e.g., the first statement of [1, Theorem A]) that $\zeta = \omega - \omega_P$ is an exact $(1,1)$-form, i.e., $\zeta = d\xi$ for some real $1$-form $\xi$ on $\mathbb{C}^m \times Y$.
My question concerns obtaining a more direct proof of Statement (ii) of [1, Theorem A]. That is, I am attempting to prove that there exists an automorphism $T$ of $\mathbb{C}^m$ such that $$\omega = T^{\ast} \omega_{\mathbb{C}^m} + \omega_Y + \sqrt{-1}\partial \overline{\partial} \varphi,$$ for some smooth $\mathbb{R}$-valued function $\varphi$.
The proof given in [1, Proposition 3.1] is too abstract, and I would like a proof along the same lines as the Poincaré lemma and the Dolbeault lemma. In [2], the authors write $\omega$ can be expressed in local coordinates as \begin{eqnarray*} \omega &=& \hat{\omega}_{\mathbb{C}^m} + \omega_Y + \frac{1}{2} \sum_{i=1}^n dz^i \wedge \eta^i + \frac{1}{2} \sum_{i=1}^n d\overline{z}^i \wedge \overline{\eta}^i, \end{eqnarray*}
where $\eta^i = \left( \frac{\partial}{\partial z^i} \ \llcorner \ \omega \right) $ (here $\llcorner$ denotes the interior product), and $\hat{\omega}_{\mathbb{C}^m} = \frac{1}{2} \sum_{i,j=1}^m u_{i \overline{j}} dz^i \wedge d\overline{z}^j$ for some Hermitian matrix $(u_{i \overline{j}})$.
I might be able to make a lot of progress if $\sum_{i=1}^m dz^i \wedge \eta^i + \sum_{i=1}^m d\overline{z}^i \wedge \overline{\eta}^i$ could be written as $\partial \overline{\partial}\varphi$ for some smooth function $\varphi$.
References:
[1] Hein, H.-J., A Liouville theorem for the complex Monge-Ampère equation on product manifolds, arxiv: 1701.05147. (https://arxiv.org/abs/1701.05147)
[2] Li, C., Li, J., Zhang, X., A mean value formula and a Liouville theorem for the complex Monge-Ampère equation, arxiv: 1709.05754. (https://arxiv.org/abs/1709.05754)
An attempt: Let $(z_1, ..., z_m, z_{m+1}, ..., z_{m+n})$ denote the local coordinates on $\mathbb{C}^m \times Y$. Since the Kähler forms $\omega$ and $\omega_{\mathbb{C}^m} + \omega_Y$ are cohomologous, there exists a real $1$-form $\xi$ such that $$\omega = \omega_{\mathbb{C}^m} + \omega_Y + d \xi.$$ Comparing the degrees of these forms, it is clear that if we decompose $\xi = \xi^{1,0} + \xi^{0,1}$ according the decompositon $\Lambda^1 = \Lambda^{1,0} \oplus \Lambda^{0,1}$, then $\partial \xi^{1,0} = 0 = \overline{\partial} \xi^{0,1}$. Hence, we see that \begin{eqnarray*} \omega &=& \omega_{\mathbb{C}^m} + \omega_Y + \partial \xi^{0,1} + \overline{\partial} \xi^{1,0}. \end{eqnarray*}
Following [2], set $\eta^i = \left( \frac{\partial}{\partial z^i} \ \llcorner \ \omega \right) \Bigg \vert_{\{ z \} \times Y}$. If we write $\omega = \frac{\sqrt{-1}}{2} \sum_{i,j=1}^{m+n} g_{i \overline{j}} dz^i \wedge d\overline{z}^j$, then \begin{eqnarray*} \eta^i &=& \frac{\sqrt{-1}}{2} \sum_{j=m+1}^{m+n} g_{i \overline{j}} d\overline{z}^j. \end{eqnarray*}
Hence, we see that \begin{eqnarray*} &&\sum_{i=1}^n dz^i \wedge \eta^i + \sum_{i=1}^n d\overline{z}^i \wedge \overline{\eta}^i \\ &=& \frac{\sqrt{-1}}{2} \sum_{i=1}^n \sum_{j=m+1}^{m+n} g_{i \overline{j}} dz^i \wedge d\overline{z}^j+ \frac{\sqrt{-1}}{2} \sum_{i=1}^n \sum_{j=m+1}^{m+n} g_{j \overline{i}} dz^j \wedge d\overline{z}^i. \end{eqnarray*}
The claim from [2] is that there now exists a Hermitian matrix $(u_{i \overline{j}})$ such that $\omega = \hat{\omega}_{\mathbb{C}^m} + \omega_Y + \frac{1}{2} \sum_{i=1}^n dz^i \wedge \eta^i + \frac{1}{2} \sum_{i=1}^n d\overline{z}^i \wedge \overline{\eta}^i$.
Here is a partial answer to the above question.
Write $\zeta = \omega - (\omega_{\mathbb{C}^m} + \omega_Y)$, which we know to be exact, so there is a $1$-form $\xi$ such that $\zeta = d\xi$. Of course, we observe that \begin{eqnarray*} \zeta = d \xi = \partial \xi^{0,1} + \overline{\partial} \xi^{1,0}. \end{eqnarray*} Let $(z^1, ..., z^m, z^{m+1}, ..., z^{m+n})$ denote the local holomorphic coordinates on $\mathbb{C}^m \times Y$. Then \begin{eqnarray*} \zeta &=& \sum_{j=1}^{m+n} \frac{\partial \xi^{0,1}}{\partial z^j} dz^j + \sum_{j=1}^{m+n} \frac{\partial \xi^{1,0}}{\partial \overline{z}^j} d\overline{z}^j. \end{eqnarray*}
So far, this is obvious. Now, \begin{eqnarray*} \frac{\partial}{\partial z^j} \ \llcorner \ \zeta &=& \frac{\partial \xi^{0,1}}{\partial z^j} \end{eqnarray*}
because the set of $\dfrac{\partial}{\partial z^k}$ is a dual basis to $dz^k$, i.e., $dz^i \left( \dfrac{\partial}{\partial z^j} \right) = \delta_j^i$, where $\delta_j^i$ is the Kronecker-Delta. Hence, it is clear that $$\left( \frac{\partial}{\partial z^j} \ \llcorner \ \zeta \right) \Bigg \vert_{\{z \} \times Y} = \frac{\partial \xi^{0,1}}{\partial z^j} \Bigg \vert_{\{ z \} \times Y}.$$
Now recall that $f(z) = \Phi[\xi^{0,1}](z) = [\xi^{0,1} \vert_{\{ z \} \times Y}]$, therefore, \begin{eqnarray*} \frac{\partial f}{\partial z^j} = \left[ \frac{\partial \xi^{0,1}}{\partial z^j} \Bigg \vert_{\{ z \} \times Y} \right], \end{eqnarray*}
so it seems to work. I'll touch up the details later.