I cannot understand why the inner product involves conjugate in the second term and where does the $\partial_{x_j}^2$ come from and why it can be moved into the conjugate? Isn't partial derivative tied to the variable you are differentiaing?
Is there nicer proof than this?
$\Delta f =-\operatorname{div}(\operatorname{grad}(f))$ so using the identity $ \operatorname{div}(f F) =\langle \operatorname{grad}f, F \rangle +f \operatorname{div} F$, you get that $\int_M \Delta f \psi dV = \int_M-\operatorname{div}(\operatorname{grad}(f))\psi dV = \int_M \operatorname{div(\psi \operatorname{grad}}(-f))dV + \int_M \langle \operatorname{grad}\psi, \operatorname{grad}(f) \rangle dV$
By Stokes $\int_M \operatorname{div(\psi \operatorname{grad}}(-f))dV=0$ if $M$ has no boundary. And repeating the same steps on $\psi $ instead of $f$ you get that $\int_M \langle \operatorname{grad}\psi, \operatorname{grad}(f) \rangle dV=\int_M \Delta \psi\ fdV$.