$$ F(Du(x+h) - F(Du(x)) = \int_0^1 \int_0^t \frac{\partial ^2}{\partial s\partial s}F(sDu(x+h) + (1-s)Du(x)) \ ds dt = \int_{0}^{1} \frac{\partial}{\partial t}F(tDu(x+h) + (1-t)Du(x)) \ dt $$
I am not sure to can put $s$ on the limits of integration, but maybe it can be correct. I know that I could introduce another variable $t$, but I need the formula in this way.
Do you mean something like $$ \begin{split} I &= \int_0^1 \int_0^s \frac{\partial ^2}{\partial s\partial s}F(sDu(x+h) + (1-s)Du(x)) ds ds \\ &= \int_0^1 \left[ \int_0^s \frac{\partial}{\partial s} \left[\frac{\partial}{\partial s} F(sDu(x+h) + (1-s)Du(x)) \right] ds \right] ds \\ &= \int_0^1 \frac{\partial}{\partial s} F(sDu(x+h) + (1-s)Du(x)) ds\\ &= F(sDu(x+h) + (1-s)Du(x)) \bigg|_{s=0}^{s=1} \\ &= F(Du(x+h)) - F(Du(x)) \end{split} $$