The equation for the upper half of the hyperboloid is the following, where $a$, $b$, and $c$ are all positive real numbers
$$z=\sqrt{\frac{x^{2}}{a^2}+\frac{y^{2}}{b^2}+c^2}$$
The radius of the sphere is the reciprocal of its principal curvature, which is the same in all directions on the sphere. Consider a circle which has the same radius as the sphere. That circle's curvature is the same value as the principal curvature of the sphere. To solve for that circle's curvature, find the curvature at $(0,0,c)$ of the cross-sectional slice with the plane $x=0$ or $y=0$, depending on whether $a$ or $b$ is larger.
For instance, if $a>b$ consider the cross section $z=\sqrt{\frac{y^2}{b^2}+c^2}$. The curvature for this curve in terms of y can then easily be derived, and our desired curvature for the osculating circle of this curve is obtained by plugging in the $y$ value zero.
After solving for the curvature of the cross-section at $(0,0,c)$, it follows that:
If $a>b$ the osculating sphere kissing the hyperboloid at the point $(0,0,c)$ can be parametrized by
$$g(u,v)=(b^{2}c\cos\left(v\right)\sin (u),b^{2}c\sin\left(v\right)\sin (u),b^{2}c\cos (u)+\left(b^{2}+1\right)c$$
If $a<b$ the osculating sphere kissing the hyperboloid at the point $(0,0,c)$ can be parametrized by the same equations as above except with every $b$ switched out for an $a$
Here I have graphs for when $a>b$