Using Green's formula we will prove that
$\int\limits_{\partial D} \frac{\partial u}{\partial n} \,dS = \iint\limits_D \Delta u \,dx\,dy.$
Let $D$ be the region for which Green's theorem holds and $u: \mathbb{R}^2 \rightarrow \mathbb{R}$. Let us denote the unit external tangent vector as $n$, i.e. if $\sigma(t)=(\sigma_1(t),\sigma_2(t))$ is a parameterization of the boundary $\partial D$ consistent with Green's theorem, then $${n}=\dfrac{(\sigma_2^{'}(t),\sigma_1^{'}(t))}{||\sigma^{'}(t)||},$$ where $||\sigma^{'}(t)||=\sqrt{\sigma_1^{'}(t)^2+\sigma_2^{'}(t)^2}$,
$\sigma:(a,b)\rightarrow \partial D.$
Let's calculate the directional derivative $$\left.\frac{d}{dS}u((x,y)+s_n)\right|_{s=0}=\frac{\partial u}{\partial x}\cdot n_1 + \frac{\partial u}{\partial y}\cdot n_2=\langle\nabla u,n\rangle,$$ where $n=(n_1,n_2)$ and $\nabla u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y})$. We have that $$\langle\nabla u,n\rangle=\frac{\partial u}{\partial x}\cdot \frac{\sigma_2^{'}(t)}{||\sigma^{'}(t)||} + \frac{\partial u}{\partial y}\cdot \frac{-\sigma_1^{'}(t)}{||\sigma^{'}(t)||},$$ so $$\int\limits_{\partial D} \langle\nabla u,n\rangle \, dS = \int\limits_a^b \left(\frac{\partial u}{\partial x}(\sigma(t))\cdot \frac{\sigma_2^{'}(t)}{||\sigma^{'}(t)||} + \frac{\partial u}{\partial y}(\sigma(t))\cdot \frac{-\sigma_1^{'}(t)}{||\sigma^{'}(t)||} \right)\cdot ||\sigma^{'}(t)|| \, dt=$$ $$=\int\limits_a^b \left(\frac{\partial u}{\partial x}(\sigma(t)) {\sigma_2^{'}(t)} - \frac{\partial u}{\partial y}(\sigma(t)){\sigma_1^{'}(t)} \right) \,dt= \int\limits_{\partial D} \frac{\partial u}{\partial x}\,dy - \frac{\partial u}{\partial y}\,dx\,.$$ Applying Green's theorem for the last integral, we obtain: $$\int\limits_{\partial D} \langle\nabla u,n\rangle \, dS =\int\limits_D \left( \frac{\partial^2 u}{\partial y} + \frac{\partial^2 u}{\partial x}\right)\,dx\,dy\,.$$
How to similarly prove another equality?
$\iint\limits_D v\Delta u \,dx\,dy=-\iint\limits_D \left(\frac{\partial u}{\partial x}\cdot \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\cdot \frac{\partial v}{\partial y} \right) \,dx\,dy + \int\limits_{\partial D} v\frac{\partial u}{\partial n}\, dS$
Note that $v\nabla^2u=\nabla \cdot (v\nabla u)-\nabla u\cdot \nabla v$. Hence, we have
$$\begin{align} \iint_D v\nabla^2u\,dx\,dy&=\iint_D \left(\nabla \cdot (v\nabla u)-\nabla u\cdot \nabla v\right)\,dx\,dy\\\\ &=\iint_D \nabla \cdot (v\nabla u)\,dx\,dy-\iint_D \nabla u\cdot \nabla v\,dx\,dy\\\\ &=-\iint_D \nabla u\cdot \nabla v\,dx\,dy+\oint_{\partial D}\hat vn\cdot \nabla u\,ds\\\\ &=-\iint_D \left(\frac{\partial u}{\partial x}\cdot \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\cdot \frac{\partial v}{\partial y}\right)\,dx\,dy+\oint_{\partial D}v\frac{\partial u}{\partial n}\,ds \end{align}$$
as was to be shown!