Point $(b')$ little bit confuses me. Let $\nabla \cdot \mathbf{F}=0$ then $d\omega_{\mathbf{F}}=0$. We see that $\mathbf{F}\in C'$. Since $\omega_{\mathbf{F}}$ is closed in $E$ then by theorem 10.40 it is exact, i.e. exists $1$-form $\lambda$ such that $d\lambda=\omega_{\mathbf{F}}$. Also $\lambda$ assosiates to some vector field $\mathbf{G}$ in $E$, i.e. $d\lambda_{\mathbf{G}}=\omega_{\mathbf{F}}$. Hence $\mathbf{F}=\nabla\times \mathbf{G}$.
Remark: From $d\lambda_{\mathbf{G}}=\omega_{\mathbf{F}}$ we get: $$F_1=D_2G_3-D_3G_2\in C',$$ $$F_2=D_3G_1-D_1G_3\in C',$$ $$F_3=D_1G_2-D_2G_1\in C',$$ How he conclude that $G\in C''$? Because from above equalities does not follow that $G\in C''$.