Riemannian connections: how to understand $\overline{\triangledown}_X \langle Y,Z \rangle$

I have a question regarding a comment in Lee's book "Riemannian Geometry - an Introduction to Curvature".

On page 52, Lee introduces the Euclidean Connection as the map $\overline{\triangledown} : \mathcal{T}(M) \times \mathcal{T}(M) \to \mathcal{T}(M)$ that acts locally on a vector field $Y$ by $$\overline{\triangledown}(X,Y) = \overline{\triangledown}_X (Y_i \partial_i):= (XY^i) \partial_i$$

But then, on page 67 he makes the following remark:

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

The thing that confuses me a little is I don't know what the left hand means, as I understood the connection is defined on $\mathcal{T}(M)$, but $\langle Y,Z \rangle \notin \mathcal{T}(M)$. How can I understand the left hand side of the above equation?

One guess that I have is that I should understand it in terms of the connection on the tensor bundle $T^2(M)$, as defined in Lemma 4.6 (page 53) , but then I have

Think of $<Y,Z>$ as being a function defined on $\mathbb R\times \mathbb R$, and think of $\overline \nabla_X$ as being the directional derivative w.r.t. $X$. –  treble Apr 4 '12 at 23:53
Yuri is right. That is, one can talk about $\overline{\nabla}_Xf := Xf$ where $f$ is a smooth function, and in this case $f = \langle Y, Z \rangle$. –  Jesse Madnick Apr 5 '12 at 16:35