$M\subset \mathbb R^{n+1}$ is a n-dimensional surface. $\nu$ is normal vector of $M$. $f$ is a function on $M$. $\nabla $ is Euclidean connection. How to show $\langle\nabla_\nu(\nabla f),\nu\rangle=\nabla^2f(\nu,\nu)$ ?
What I try: $$ \nabla ^2 f(\nu,\nu) = \nabla _\nu\nabla _\nu f =\nabla _\nu \langle\nabla f , \nu\rangle =\langle \nabla_\nu (\nabla f ), \nu \rangle + \langle \nabla f , \nabla _\nu \nu \rangle $$
But , how to show $\langle \nabla f , \nabla _\nu \nu \rangle =0$ ?
Your first equality is incorrect - the second covariant derivative is $$\nabla^2 f(X,Y) = (\nabla_X (\nabla f))(Y)=\nabla_X ((\nabla f)(Y)) - (\nabla f)(\nabla_X Y)=\nabla_X\nabla_Y f - \nabla_{\nabla_X Y} f.$$