About the differentiation formula of $u(t)$: Since \begin{align} u(t)&=e^{-\int_{t_0}^ta(s)\mathrm{d}s}u(t_0)+\int_{t_0}^t\sigma(v)e^{-\int_{s}^ta(s)\mathrm{d}s}\mathrm{d}Z(v)\\ &=e^{-\int_{t_0}^ta(s)\mathrm{d}s}\Bigl[u(t_0)+\int_{t_0}^t\sigma(v)e^{\int_{t_0}^va(s)\mathrm{d}s}\mathrm{d}Z(v)\Big]\stackrel{\text{def}}{=}x(t)y(t).\\ \mathrm{d}x(t)&=-a(t)e^{-\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}t=-a(t)x(t)\mathrm{d}t,\\ \mathrm{d}y(t)&=\sigma(t)e^{\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}Z(t)=\sigma(t)[x(t)]^{-1}\mathrm{d}Z(t). \end{align} Therefore, $$ du(t)=y(t)\mathrm{d}x(t)+x(t)\mathrm{d}y(t) =-a(t)u(t)\mathrm{d}t+\sigma(t)\mathrm{d}Z(t). $$

About the differentiation formula of $u(t)$: Since \begin{align} u(t)&=e^{-\int_{t_0}^ta(s)\mathrm{d}s}u(t_0)+\int_{t_0}^t\sigma(v)e^{-\int_{v}^ta(s)\mathrm{d}s}\mathrm{d}Z(v)\\ &=e^{-\int_{t_0}^ta(s)\mathrm{d}s}\Bigl[u(t_0)+\int_{t_0}^t\sigma(v)e^{\int_{t_0}^va(s)\mathrm{d}s}\mathrm{d}Z(v)\Big]\stackrel{\text{def}}{=}x(t)y(t).\\ \mathrm{d}x(t)&=-a(t)e^{-\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}t=-a(t)x(t)\mathrm{d}t,\\ \mathrm{d}y(t)&=\sigma(t)e^{\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}Z(t)=\sigma(t)[x(t)]^{-1}\mathrm{d}Z(t). \end{align} Therefore, $$ du(t)=y(t)\mathrm{d}x(t)+x(t)\mathrm{d}y(t) =-a(t)u(t)\mathrm{d}t+\sigma(t)\mathrm{d}Z(t). $$