Constructing such a surface is not too hard.
Note that for a smooth surface, Gaussian curvature is a continuous function. Hence if you have a connected surface it suffices to have points of arbitrarily large (in absolute value) Gaussian curvatures of either sign.
So consider the graph of the function
$$ z = f(x,y) = \cos(2\pi x) + x y^2 $$
We have that
$$ \mathrm{d}f = (-2\pi \sin (2\pi x) + y^2)\mathrm{d}x + 2xy \mathrm{d}y $$
and in particular
$$ \mathrm{d}f(n,0) = 0 $$
for $n \in \mathbb{Z}$. At those points the Gaussian curvature is simply the determinant of the Hessian
$$ K(n,0) = \det\begin{pmatrix} - 4\pi^2\cos(2\pi n) & 2\cdot 0 \\ 2\cdot 0 & 2n \end{pmatrix} = - 8 \pi^2 n$$
Do you perhaps intend to add other criteria to your surface? The clothoid has the property that every curvature value is realised by exactly one point. This is of course not possible for a surface, but maybe you want a surface where every the sets $K^{-1}(k)$ are all homeomorphic or something like that?
If you want a surface that is contained in a compact set in $\mathbb{R}^3$, consider the following map:
Let $D = \{ (\theta,s)\in \mathbb{R}^2 : \theta\in (-\pi,\pi), |s| < 1, |s^2 \tan \theta| < \frac12\}$
Let $\phi:D\to \mathbb{R}\times\mathbb{R}_+ \times\mathbb{S}^1$, the cylindrical coordinate representation of $\mathbb{R}^3$, be given by
$$\phi(\theta,s) = (s,\frac12 \tan\theta s^2,\theta) $$
The principle curvatures at point $(0,s)$ are $\{ 1, -\tan\theta\}$ and so the Gauss curvataure is $-\tan\theta$.
Our choice of domain guarantees that $\phi$ is an embedding and that $\phi(D)$ is contained in a ball of sufficiently large radius.