Reading through Carmo's Differential Geometry of Curves and Surfaces, the following statement is made on an algebraic value for the covariant derivative (pg. 252, section 4-4):
Proposition 3: Let $x(u,v)$ be an orthogonal parametrization (that is, $F = 0$), of a neighborhood of an oriented surface $S$, and $w(t)$ be a differentiable field of unit vectors along the curve $x(u(t), v(t))$. Then, $$[\frac{Dw}{dt}] = \frac{1}{2\sqrt{EG}}\{G_{u}\frac{dv}{dt} - E_{v}\frac{du}{dt}\} + \frac{d \varphi}{dt},$$ where $\varphi(t)$ is the angle from $x_{u}$ to $w(t)$ in the given orientation
The proof of this statement is given in the text. My question is: what does this mean? My linear algebra is a bit rusty, so I'm not sure if that's the intuition I'm missing, but is there any extra intuition, geometric or otherwise, to be taken away from this expression of the covariant derivative (regarding the fundamental form and what it means for the surface), or is it just a tidy way of expressing the derivative that will make computation for other proofs more manageable? It certainly achieves the latter part, but I feel like I'm missing some important details on the former, more intuitive part.
Any input is much appreciated.
You can read more about the intuition in my (free) differential geometry text (linked in my profile). It's a superposition principle: The first term is telling you at what rate the frame $\{x_u/\sqrt E,x_v/\sqrt G\}$ is rotating, and you add to that $d\phi/dt$, which tells you at what rate your unit-length vector field $w$ is rotating relative to the frame.