The first equality is simply the first green identity. But what happened in the second equality? I cannot see the usage of the divergence theorem. For the divergence theore to be applied I should have the integral of the divergence of a vector field.
The step from the first to the second inequality is simply the application of the following vector identity (where $a=a(x)$ is a scalar function and $\mathbf{b}=\mathbf{b}(x)$ is a vector function): $$ \nabla\cdot(a\mathbf{b})=a\nabla\cdot\mathbf{b}+\nabla a\cdot\mathbf{b} $$ Then putting $a=u$ and $\mathbf{b}=\nabla\Phi$ you get $$ \nabla\cdot(u\nabla\Phi)=u\nabla\cdot\nabla\Phi+\nabla u\cdot\nabla\Phi\iff \nabla\Phi\cdot\nabla u=\nabla\cdot(u\nabla\Phi)-u\nabla\cdot\nabla\Phi\tag{1}\label{1} $$ Then, using \eqref{1} in the volume integral above, obtained from the first Green identity, i.e. $$ I_\epsilon(x)=\int_{V_\epsilon}\nabla_y\Phi(y-x)\cdot\nabla_y u(y)\mathrm{d}y, $$ and applying the divergence (Gauss-Green) theorem to the first one of the two integrals you get, you obtain the sought for result.
Edit. In order to explain better my answer, I have added all the steps. Precisely, $$ \begin{split} I_\epsilon(x)&=\int_{V_\epsilon}\nabla_y\Phi(y-x)\cdot\nabla_y u(y)\mathrm{d}y\\ &=\int_{V_\epsilon}\Big[\nabla_y\cdot\big(u(y)\nabla_y\Phi(y-x)\big)-u(y)\nabla_y\cdot\nabla_y\Phi(y-x)\Big]\mathrm{d}y\\ &=\int_{V_\epsilon}\nabla_y\cdot\big(u(y)\nabla_y\Phi(x-y)\big)\mathrm{d}y- \int_{V_\epsilon}u(y)\Delta_y\Phi(y-x)\mathrm{d}y\end{split} $$ Now the first integral at the second side of the is the integral of the divergence of the vector field $y\mapsto u(y)\nabla_y\Phi(x-y)$, and applying to it the divergence (Gauss-Green) theorem we get $$ \begin{split} I_\epsilon(x)&=\int_{\partial V_\epsilon}u(y)\nabla_y\Phi(x-y)\cdot\nu_y\mathrm{d}S_y- \int_{V_\epsilon}u(y)\Delta_y\Phi(y-x)\mathrm{d}y\\ &=\int_{\partial V_\epsilon}\frac{\partial\Phi}{\partial\nu_y}(y-x)u(y)\mathrm{d}S_y- \int_{V_\epsilon}\Delta_y\Phi(y-x)u(y)\mathrm{d}y \end{split} $$