Let $X_{t}$ be the Hamiltonian vector field along $L_{t}$ associated to $h_{t}$, namely $dh_{t}=\omega(X_{t},\cdot)$. We calculate the time derivative of $f_{L_{t}}$: along $L_{t}$

$$\mathcal{L}_{X}\lambda=\iota_{X}d\lambda+d(\iota_{X}\lambda)=\iota_{X}\omega+d(\iota_{X}\lambda)=d(h_{t}+\iota_{X}\lambda),$$

so there is a preferred parallel transport of $f_{L}$ along the path $L_{t}$,

$$\partial_{t}f_{L_{t}}=h_{t}+\iota_{X}\lambda.$$

Hence

$$\partial_{t}\int_{L_{t}}f_{L_{t}}\text{Im}(e^{-i\hat{\theta}}\Omega)=\int_{L_{t}}(h_{t}+\iota_{X}\lambda)\text{Im}(e^{-i\hat{\theta}}\Omega)+\int_{L_{t}}f_{L_{t}}\mathcal{L}_{X}\text{Im}(e^{-i\hat{\theta}}\Omega).$$

Now by the Cartan formula and the closedness of $\Omega$,

$$\mathcal{L}_{X}\text{Im}(e^{-i\hat{\theta}}\Omega)=d\iota_{X}\text{Im}(e^{-i\hat{\theta}}\Omega),$$

so after integration by part,

$$\int_{L_{t}}f_{L_{t}}\mathcal{L}_{X}\text{Im}(e^{-i\hat{\theta}}\Omega)=-\int_{L_{t}}df_{t}\wedge\iota_{X}\text{Im}(e^{-i\hat{\theta}}\Omega)=-\int_{L_{t}}\lambda\wedge\iota_{X}\text{Im}(e^{-i\hat{\theta}}\Omega).$$

Combining the above,

$$\partial_{t}\int_{L_{t}}f_{L_{t}}\text{Im}(e^{-i\hat{\theta}}\Omega)=\int_{L_{t}}h_{t}\text{Im}(e^{-i\hat{\theta}}\Omega)+\int_{L_{t}}\iota_{X}(\lambda\wedge\text{Im}(e^{-i\hat{\theta}}\Omega)).$$

Integrating in $t$ gives the result. ∎