In Riemannian manifold, let $\{e_1, \cdots e_n\}$ be an orthonormal frame. Then there is a dual coframe $\{w^1, \cdots, w^n\}$.
$f_i := \nabla f( e_i)$
$f_{ij} : = \nabla^2f(e_i, e_j)$
$f_{ijk} : = \nabla^3 f(e_i, e_j, e_k)$
$\vdots$
I want to show that $f_i = e_i(f)$ and $f_{ij} = e_j(f_i)$. The first identity is trivial: $$f_i = \nabla f( e_i) = e_i(f)$$ To show that the second identity holds, I tried as below. \begin{align*} f_{ij}=f_{ji} & = \nabla^2 f (e_j ,e_i)\\ & = \nabla_{e_j,e_i}^2 f\\ & = \nabla_{e_j} \nabla_{e_i}f - \nabla_{\nabla_{e_j}e_i}f\\ &=(f_i)_j - \nabla_{\nabla_{e_j}e_i}f \end{align*} I want to show that $\nabla_{\nabla_{e_j}e_i}f = 0$. How can I show this?