A question about the definition of $\overline{\nabla}_X \langle Y,Z\rangle$ Lee's "Riemannian Manifolds" has the following theorem:

The Euclidean connection on $\Bbb{R}^n$ has one very nice property with respect to the Euclidean metric: it satisfies the product rule
  $$\overline{\nabla}_X \langle Y,Z\rangle=\langle \overline{\nabla}_XY,Z\rangle + \langle Y,\overline{\nabla}_XZ\rangle$$

But $\langle Y,Z\rangle$ is not a vector field. How does $\overline{\nabla}_X \langle Y,Z\rangle$ make sense?
 A: Yes, $\langle Y, Z \rangle$ is a scalar, not a vector, field.  $\overline \nabla_X \langle Y, Z \rangle$ make sense by virtue of the fact that, for scalar fields $f$, we define
$\overline \nabla_X f = X[f] = \dfrac{\partial f}{\partial t}, \tag 1$
where $t$ is a parameter along the integral curves of $X$:
$X = \dfrac{\partial}{\partial t}; \tag 2$
then the formula
$\overline \nabla_X \langle Y, Z \rangle = \langle \overline \nabla_X Y, Z \rangle + \langle Y, \overline \nabla_X Z \rangle \tag 3$
expresses
$\overline \nabla_X \langle Y, Z \rangle = X[\langle Y, Z \rangle] \tag 4$
as the sum of the two inner products $\langle \overline \nabla_X Y, Z \rangle$ and $\langle Y, \overline \nabla _X Z \rangle$.  
This is all of course just a generalization of the well-known formula in Euclidean spaces:
$\dfrac{d}{dt} (\vec v(t) \cdot \vec y(t)) = \dot {\vec v}(t) \cdot \vec y(t) + \vec v(t) \cdot \dot {\vec y}(t), \tag 5$
where $t$ is the parameter along some curve on which $\vec v(t)$, $\vec y(t)$ are defined.
