We need to calculate the expansion for $dw_{1}\wedge\ldots\wedge dw_{n-1}$. First, by Lemma 3.27,

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$$dw_{1}=a_{1}(dy+yd_{H}\log a_{1})+\widetilde{O}(|y|)dy+\widetilde{O}(|y|^{2}),$$

where $a_{1}=|\sigma|^{-1}$. Notice that

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$$d_{H}\log a_{1}=2\partial_{H}\log a_{1}-\sqrt{-1}d_{H}^{c}\log a_{1}.$$

Applying Lemma 3.25,

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$$d_{H}\log a_{1}=2(\partial_{H}\log a_{1}+\sqrt{-1}\Gamma).$$

Next, applying Lemma 3.27 to $w_{j}$’s for $j\geq 2$,

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$$dw_{j}=dw_{j}^{\prime}+c_{j}dy+ydc_{j}+\widetilde{O}(|y|)dy+\widetilde{O}(|y|^{2}).$$

Since it holds that

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$$\partial_{H}\log a_{1}\wedge\Omega_{H}\equiv 0,$$

then taking the wedge product,

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$$dw_{1}\wedge\cdots\wedge dw_{n-1}=a_{1}(dy+2\sqrt{-1}y\Gamma)\wedge\Omega_{H}+\widetilde{O}(|y|)dy+\widetilde{O}(|y|^{2}).$$

On the other hand, we have the expansion of $f$,

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$$f=f|_{H}+\frac{\partial f}{\partial y}|_{H}\cdot y+\frac{\partial f}{\partial\bar{y}}|_{H}\cdot\bar{y}+\widetilde{O}(|y|^{2}).$$

Therefore,

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$$\Omega_{D}=f|_{H}a_{1}(dy+2\sqrt{-1}y\Gamma)\wedge\Omega_{H}+\widetilde{O}(|y|)dy+\widetilde{O}(|y|^{2}).$$

So we obtain the conclusion by taking $F=f|_{H}\cdot a_{1}$.

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