Coordinate-free differentiation techniques in Riemannian geometry I encountered the following identities while reading this article on global calculus (p. 10): $$ d(\|df\|^2)=2\mathop{\iota_{\mathop{\mathrm{grad}} f}} \mathop{\mathrm{Hess}} f, $$ $$ \mathop{\mathrm{grad}}(\|df\|^2)=2\mathop{\nabla_{\mathop{\mathrm{grad}} f}}\mathop{\mathrm{grad}} f $$ Here $\mathop{\mathrm{Hess}} f = \nabla d f$ is the covariant derivative of the 1-form $df$, and the norm $\|\cdot\|$ is given by Riemannian metric: $\|\upsilon\|^2=g(\upsilon,\upsilon)$. I wonder how does one usually derive these identities (without using coordinates)?

By simply using the definition. The first thing to remember is that the covariant derivative commutes with metric and inverse metric. The second thing to remember is that the covariant and exterior derivatives agree when acting on a scalar function. So,

$$ d(\|df\|^2) = \
abla(\|df\|^2) = \
abla g^{-1}(df,df) = g^{-1}(\
abla df, df) + g^{-1}(df, \
abla df) $$

Next you use the symmetry of $g^{-1}$ as a bilinear form, and you use the definition for raising indices that for any one form $\omega$,

$$ \iota_{\mathrm{grad}~f} = \omega(\mathrm{grad}~f) = g^{-1}(\omega, df) $$

to arrive at

$$ d(\|df\|)^2 = 2 \iota_{\mathrm{grad}~f} \mathrm{Hess}~f $$

The computation for the second identity is similar. (Note that for the notations to make sense, you also need to use the fact that for a scalar function $f$, the second derivatives commute. In other words, the Hessian is a symmetric (0,2) tensor.)