While it's true that the computation of $K$ is much easier in orthogonal and especially in isothermal coordinates, finding such coordinates requires solving a PDE. I don't think it would be worth the effort in this case. Your coefficients $E,F,G$ are about as simple as one could have, the partials are immediately found and the determinants are not too bad. Everything is quite doable by hand.
With apologies for using a proprietary system (Maple), here is the general Brioschi formula:
with(LinearAlgebra):
E := 1+v^2; F := 2*u*v; G := 1+u^2;
A := Matrix([[-diff(E,v,v)/2+diff(F,u,v)-diff(G,u,u)/2, diff(E,u)/2, diff(F,u)-diff(E,v)/2], [diff(F,v)-diff(G,u)/2, E, F], [diff(G,v)/2, F, G]]);
B := Matrix([[0, diff(E,v)/2, diff(G,u)/2], [diff(E,v)/2, E, F], [diff(G,u)/2, F, G]]);
K := simplify((Determinant(A)-Determinant(B))/(E*G-F^2)^2);
Output: $$K=\frac{u^2+v^2}{(1+u^2+v^2-3u^2v^2)^2}$$
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$If you have a surface embedded in $\Reals^{3}$, you can (as you note) use the "ambient" Euclidean inner product to take dot products of tangent vectors. However, that's mild overkill; in order to take dot products of tangent vectors to a surface, all you really need is the first fundamental form.
If
$$
E = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd u}\right\rangle,\quad
F = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd v}\right\rangle,\quad
G = \left\langle\frac{\dd q}{\dd v}, \frac{\dd q}{\dd v}\right\rangle,
$$
and if
$$
v = a_{1}\, \frac{\dd q}{\dd u} + a_{2}\, \frac{\dd q}{\dd v},\qquad
w = b_{1}\, \frac{\dd q}{\dd u} + b_{2}\, \frac{\dd q}{\dd v},
$$
i.e., if $v = (a_{1}, a_{2})$ and $w = (b_{1}, b_{2})$ with respect to the coordinate basis fields, then
$$
\langle v, w\rangle
= a_{1} b_{1} E + (a_{1} b_{2} + a_{2} b_{1}) F + a_{2} b_{2} G.
$$
That's the content of the first paragraph of your excerpt. The second paragraph asserts that $\langle v, w\rangle$ is independent of the choice of local coordinates. This is in some sense "obvious" if you know you have a surface embedded in $\Reals^{3}$, but it's surprising if you think of the first fundamental form as "extra structure" imposed on an open set in $\Reals^{2}$, i.e., in a coordinate neighborhood.
Remarkably, there's non-trivial "intrinsic" geometry of a surface captured by its first fundamental form, even though the first fundamental form doesn't uniquely determine an embedding of the surface in $\Reals^{3}$.
Best Answer
$\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$There's a specific, formally simple relationship between a coordinate system on a regular surface and the components of the first fundamental form: In customary notation, if $\Vec{x}$ is a regular parametrization, then $$ E(u, v) = \Brak{\Vec{x}_{u}, \Vec{x}_{u}},\qquad F(u, v) = \Brak{\Vec{x}_{u}, \Vec{x}_{v}},\qquad G(u, v) = \Brak{\Vec{x}_{v}, \Vec{x}_{v}}. $$ The subtle points are:
Although two surfaces sharing a tangent plane at a point $p$ have "the same first fundamental form at $p$", they do not generally have equivalent first fundamental forms in any open neighborhood of $p$;
Perhaps surprisingly, it's not true that by using the right parameterization we can get any triplet $(E, F, G)$ we want. If it were, there would be no local invariants of a surface, because one could always pick, say, $E = G = 1$ and $F = 0$.
Globally, the first fundamental form of a surface is an inner-product-valued function on a surface.
Locally, the first fundamental form of a surface is represented by an association to each coordinate system of a $2 \times 2$ matrix-valued function whose value at each point is positive-definite, and which "transforms like an inner product" under change of coordinate. The first fundamental form itself may be viewed as the resulting equivalence class of all such pairs of "coordinate system and matrix-valued function".