The starting point is this convolution $$ \frac{\partial v_0}{\partial G}(t) = \int_0^t v_G(\xi) \Psi^{(0)}_G(t-\xi) \,d\xi. $$ Applying the product rule for differentiation \begin{align*} \frac{\partial^2v_0}{\partial G^2}(t) &= \int_0^t \left\{ \frac{\partial v_G}{\partial G}(\xi) \Psi^{(0)}_G(t-\xi) + v_G(\xi) \frac{\partial \Psi^{(0)}_G}{\partial G}(t - \xi)\right\} d\xi \\&= 2 \int_0^t \left\{ \int_0^{\xi} v_G(\eta) \Psi^{(G)}_G(\xi - \eta)\,d\eta \right\} \Psi^{(0)}_G(t-\xi) \,d\xi \end{align*}
Can anyone help on how the double convolution appears? What step is needed to understand the last equality?
EDIT:
This is the source of the equality
