I am only getting a partial answer. Here is what and how I am getting: $D$ is of form $-g^{\mu\nu}(\nabla_{\mu}\nabla_{\nu}+C)$ where $C$ is some constant, given in section 1 of the paper. I will take $C$ to be $0$ (zero) in the following also I will strip away all constant, which can be stripped, from $K$ for the ease of typing: $$K(t ; x, y ; D)=\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi$$ $$(\partial_t + D)K(t ; x, y ; D)=\partial_tK+DK$$ For the first term $$\partial_tK=-\frac{n}{2t}\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi+\frac{\sigma}{2t^2}\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right) \Xi+\frac{1}{t^{n / 2}}\Delta^{1 / 2}\exp \left(-\frac{\sigma}{2 t}\right)\Big(\partial_t \Xi\Big)$$ while the second term $$DK=-\frac{1}{t^{n/2}}g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\bigg(\Big(\Delta^{1 / 2}\Xi\Big)\exp \left(-\frac{\sigma}{2 t}\right) \bigg)$$ $$=-\frac{g^{\mu\nu}}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\nabla_{\nu}\Big(\Delta^{1 / 2}\Xi\Big)+2\nabla_{\mu}\bigg[\exp \left(-\frac{\sigma}{2 t}\right)\bigg]\nabla_{\nu}\Big(\Delta^{1 / 2}\Xi\Big)+\Big(\Delta^{1 / 2}\Xi\Big)\nabla_{\mu}\nabla_{\nu}\exp \left(-\frac{\sigma}{2 t}\right)\Bigg)$$
$$=\frac{1}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)D\Big(\Delta^{1 / 2}\Xi\Big)- 2\bigg[-\frac{1}{2t}\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\sigma\bigg]\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)+\Big(\Delta^{1 / 2}\Xi\Big)D\exp \left(-\frac{\sigma}{2 t}\right)\Bigg)$$ $$=\frac{1}{t^{n/2}}\Bigg(\exp \left(-\frac{\sigma}{2 t}\right)D\Big(\Delta^{1 / 2}\Xi\Big)+\frac{1}{t}\bigg[\exp \left(-\frac{\sigma}{2 t}\right)\nabla_{\mu}\sigma\bigg]\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)-\frac{1}{2t}\Big(\Delta^{1 / 2}\Xi\Big)\exp \left(-\frac{\sigma}{2 t}\right)D\sigma\Bigg)$$ The second term can be expanded using $\nabla^{\mu}\Big(\Delta^{1 / 2}\Xi\Big)=\Xi\nabla^{\mu}\Delta^{1 / 2}+\Delta^{1 / 2}\nabla^{\mu}\Xi$. 818