= Solution
Let $\mathbf e_r,\mathbf e_\phi,\mathbf e_z$ be the cylindrical unit vectors. The surface has an upward <normal vector> proportional to $\mathbf e_z-h'(r)\mathbf e_r$. If $N_z$ is the vertical component of the <normal force>, its radial component is $-h'(r)N_z$. In the stated approximation, vertical acceleration and the vertical driving-force component are neglected, so vertical balance gives $N_z\simeq Mg$. The horizontal contact <force> is consequently $-Mg h'(r)\mathbf e_r$. The projected velocity and its speed are
$$
\mathbf v_h=\dot r\,\mathbf e_r+r\dot\phi\,\mathbf e_\phi,\qquad
u=\sqrt{\dot r^2+r^2\dot\phi^2}.
$$
The driving <force> is parallel to this velocity, with components $F\dot r/u$ and $Fr\dot\phi/u$. Use the <acceleration in polar coordinates> and <Newton's second law> to obtain
$$
\boxed{\ddot r-r\dot\phi^2=-g h'(r)+\frac FM\frac{\dot r}{u},\qquad
r\ddot\phi+2\dot r\dot\phi=\frac FM\frac{r\dot\phi}{u}.}
$$
These equations apply where $u\ne0$; at rest the direction of an exactly velocity-aligned drive is not determined.
For a prescribed path, write $r'=dr/d\phi$, $r''=d^2r/d\phi^2$. The <chain rule> gives $\dot r=r'\dot\phi$ and $\ddot r=r''\dot\phi^2+r'\ddot\phi$. Subtract $r'/r$ times the azimuthal equation from the radial equation. The driving <force> and $\ddot\phi$ both cancel:
$$
\left(r''-r-\frac{2r'^2}{r}\right)\dot\phi^2=-g h'(r).
$$
Thus the constraint for <tangentially driven motion on an axisymmetric surface> is
$$
\boxed{\dot\phi^2=\frac{g h'(r)}{r+2r'^2/r-r''}.}
$$
The denominator must give a nonnegative squared speed, and the calculation uses $r>0$. At the polar origin a limiting Cartesian description is appropriate.
For the paraboloid and <Archimedean spiral>, put
$$
\alpha=\sqrt{\frac{2g}{\ell}},\qquad
h'(r)=\frac{2r}{\ell},\qquad r=\epsilon\phi,\quad r'=\epsilon,\quad r''=0.
$$
Then, for $\phi>0$,
$$
\dot\phi^2=\alpha^2\frac{\phi^2}{\phi^2+2}.
$$
The branch tending to $\phi=\infty$ has positive <angular velocity>, hence
$$
\dot\phi=\frac{\alpha\phi}{\sqrt{\phi^2+2}}\longrightarrow\alpha,\qquad
\ddot\phi=\frac{2\alpha^2\phi}{(\phi^2+2)^2}.
$$
Since $u=\epsilon\dot\phi\sqrt{1+\phi^2}$, solve the azimuthal equation for the required <force>:
$$
\frac FM=\epsilon\sqrt{1+\phi^2}
\left(\ddot\phi+\frac{2\dot\phi^2}{\phi}\right)
=\frac{2\epsilon\alpha^2\phi(\phi^2+3)\sqrt{\phi^2+1}}{(\phi^2+2)^2}.
$$
This is positive for $\phi>0$, consistent with a forward drive. Taking the limit gives
$$
\boxed{\dot\phi(t)\longrightarrow\sqrt{\frac{2g}{\ell}},\qquad
F(t)\longrightarrow\frac{4\epsilon Mg}{\ell}.}
$$
These are the large-time limits of the printed projected equations. Their mechanical approximation has a precise range limitation: on the exact surface, $z=h(r)$ gives $\dot z=(2r/\ell)\dot r$, which along this solution grows like $2\epsilon^2\alpha\phi/\ell$. At fixed positive $\epsilon$, it is not small as $\phi\to\infty$. Thus the requested limits are formal limits of that reduced model; neglect of vertical motion is not a uniform approximation to the full three-dimensional motion all the way to infinity.
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