Recall that thanks to Lemma 4.7, on $B\times\mathbb{C}^{n-m}$ we can write

$$\lambda_{t}^{*}p^{*}T_{-\sigma}^{*}\tilde{\omega}_{t}-p^{*}(\omega_{SF}+f^{*}\omega)=E_{t},$$

where the error term $E_{t}$ is a $(1,1)$-form that goes to zero smoothly on compact sets. From (4.8) we also have that

$$E_{t}=\lambda_{t}^{*}p^{*}(T_{-\sigma}^{*}\tilde{\omega}_{t}-f^{*}\omega-t\omega_{SF}).$$

If we restrict the form $T_{-\sigma}^{*}\tilde{\omega}_{t}-f^{*}\omega-t\omega_{SF}$ to a fiber $M_{y}$ and divide by $t$ we get

$$\frac{E_{t}}{t}\bigg|_{\{y\}\times\mathbb{C}^{n-m}}=\lambda_{t}^{*}p^{*}\left(\frac{T_{-\sigma}^{*}\tilde{\omega}_{t}|_{M_{y}}}{t}-\omega_{SF,y}\right)$$

Pulling back this via the map $\lambda_{1/t}$ (the inverse of $\lambda_{t}$) we get

$$\frac{\lambda_{1/t}^{*}E_{t}}{t}\bigg|_{\{y\}\times\mathbb{C}^{n-m}}=p^{*}\left(\frac{T_{-\sigma}^{*}\tilde{\omega}_{t}|_{M_{y}}}{t}-\omega_{SF,y}\right).$$

Explicitly we have $\lambda_{1/t}(y,z)=(y,z\sqrt{t})$, which implies that $\lambda_{1/t}^{*}dz^{i}=\sqrt{t}dz^{i}$, and so

$$\frac{\lambda_{1/t}^{*}E_{t}}{t}\bigg|_{\{y\}\times\mathbb{C}^{n-m}}(y,z)=E_{t}\bigg|_{\{y\}\times\mathbb{C}^{n-m}}(y,z\sqrt{t}),$$

which goes to zero smoothly as $t$ approaches zero, uniformly in $y$. It follows that $\frac{T_{-\sigma}^{*}\tilde{\omega}_{t}|_{M_{y}}}{t}$ converges smoothly to $\omega_{SF,y}$, and the convergence is uniform as $y$ varies on compact sets of $N\backslash f(S)$. Pulling back via $T_{\sigma}$, and using the fact that $T_{\sigma}^{*}\omega_{SF,y}=\omega_{SF,y}$, we see that also $\frac{\tilde{\omega}_{t}|_{M_{y}}}{t}$ converges smoothly to $\omega_{SF,y}$, as desired. ∎