Finding parametric curves on a sphere
Solution 1:
For your first question, suppose $\gamma$ is any curve. Then the norm of $\frac{\gamma}{||\gamma||}$ is $\frac{||\gamma||}{||\gamma||} = 1$, so the image of $\frac{\gamma}{||\gamma||}$ lives on the unit sphere in your space.
Your second question can be answered with the help of stereographic projection. We'll take a curve in the plane and project it onto the unit sphere.
(Image from Wikipedia.)
If we describe the plane with the polar coordinates $(R,\Theta)$, and the sphere with the coordinates $(\varphi,\theta)$, where $\varphi$ is the zenith angle and $\theta$ the azimuth, then the map from the plane to the sphere is given by
$$ \begin{align} \varphi &= 2 \arctan\left(\frac{1}{R}\right), \\ \theta &= \Theta. \end{align} $$
If you'd like, this can be converted to Cartesian coordinates by
$$ \begin{align} x &= \cos \theta \sin \varphi, \\ y &= \sin \theta \sin \varphi, \\ z &= \cos \varphi. \end{align} $$
As an example, if we take the logarithmic spiral defined in polar coordinates by $R = e^{\Theta/8}$, the Cartesian parametric curve describing its image on the sphere is
$$ \gamma(t) = \left(\begin{array}{c} \cos t \,\sin \left(2 \arctan \left(e^{-t/8}\right)\right) \\ \sin t \,\sin \left(2 \arctan \left(e^{-t/8}\right)\right) \\ \cos \left(2 \arctan \left(e^{-t/8}\right)\right) \end{array}\right). $$
Solution 2:
Do you really mean circle, rather than sphere?
A parametrization with constant rate of increase along the $z$ axis is $(x,y,z)=\left( \sqrt{1-t^2}\cos (a \pi t), \: \sqrt{1-t^2} \sin (a \pi t), \: t \right), \: 0<t<1$, where the $a$ parameter controls the number of revolutions. This is with $a=5$:
You can derive this by noting that $x \propto \cos(\cdot) $ and $y \propto \sin (\cdot)$. If $z$ is to increase in a linear fashion from -1 to +1 we can let $z=t$ for $t=-1 \dots 1$. Finally, since the spiral is to lie on the surface of a sphere $x^2+y^2+z^2=1$, so it follows that $x=\sqrt{1-t^2}\cos (a \pi t)$ and $y=\sqrt{1-t^2} \sin (a \pi t)$