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: $$K=\frac{u^2+v^2}{(1+u^2+v^2-3u^2v^2)^2}$$
They are the same thing written in different notation - the second equation would probably be clearer if written $2 H = \mathrm{tr}_g (b)$.
Because the second fundamental form is a $(0,2)$-tensor, it does not have a trace in the simplest sense. If you naively write down $\mathrm{tr}(b) = \sum_i b_{ii}$ then you have an expression that depends on the coordinate system/frame you compute it in. However, the metric $g$ gives us a way of taking the trace of any tensor in a meaningful way.
The "trace with respect to $g$" is this trace when calculated in an orthonormal frame, or equivalently $$\mathrm{tr}_g(b) := \mathrm{tr}(g^{-1} b) = \sum_{i,j} g^{ij} b_{ij}.$$ Since $g^{-1} b$ is a $(1,1)$-tensor, this trace does not depend on the orthonormal frame chosen - it is a function on the surface independent of the coordinates chosen.
Best Answer
The natural parametrization is $F(x,y)= (x,y,g(x,y))$.
So $F'_x=(1,0,g'_x) ; F'_y=(0,1, g'_y)$.
Then ${ds}^2= (1+{g'_x}^2){dx}^2$+$2g'_xg'_y dx dy+(1+{g'_y}^2 ){dy}^2$.
A normal vector is $F'_x \wedge F'_y=(-g'_x,-g'_y,1)$,
The Gauss map is $N(x,y)=-{1\over \sqrt {1+{g'_x}^2+{g'_x}^2}}( g'_x, g'_y,-1)$.
The second form is obtained while deriving $N(x,y)$, which yields an ugly formula.
The case where $g'_x(0,0)=g'_y(0,0)=0$ (ie the surface is horizontal) is nice and easy, the second form is given by a immediate computation $ g''(0,0)= -Hess(g)(0,0)$. Note the sign -. For instance the hemi-sphere $z= \sqrt {R^2-x^2-y^2}= R-{x^2+y^2\over 2R}+o(x^2+y^2)$ has positive curvature $1/R^2$, as $g''(0,0)=-{1\over R} Id$.