We know that for any space curve $\alpha:I\rightarrow {\rm I\!R}^3$ parametrized by arc length, there exists a "*local canonical form*" around $s_0\in I$:
$$
\alpha(s) = \begin{bmatrix}s-\frac{k^2 s^3}{6} + o(s^3)\\ \frac{k's^3}{6}+\frac{s^2 k}{2}+o(s^3) \\-\frac{s^3 k \tau}{6}+o(s^3)\end{bmatrix}
$$
where $k$ is the curvature and $\tau$ the torsion. If I understood correctly, this can be achieved by setting the origin on $\alpha(s_0)$ and using the Frenet frame $\lbrace t, n, b\rbrace$ as coordinate system.

(see: Differential Geometry of Curve and Surface, Do Carmo, Sec 1-6, p27)

**What is the local canonical form of a regular surface ?**

I saw, for example on these notes (Sec 2.4, p12), that the canonical form of a regular surface $\gamma:U\rightarrow{\rm I\!R}^3$ is: $$ \gamma(u,v) = \begin{bmatrix} u \\ v \\ \frac{1}{2}(\kappa_1 u^2 + \kappa_2 v^2) + o(u^2 + v^2)\end{bmatrix} $$ with $\kappa_1$ and $\kappa_2$ the two principal curvatures.

Is this true?

The coefficient multiplying the $uv$ term of the Taylor expansion is null. Why?

How to prove it?

How to find it? (I assume that taking the two principal curvature directions and the unit normal vector as coordinate system leads to this formula).