= Solution
Use the two pieces of the <surface integral> directly; this avoids introducing an unnecessary second <derivative> of $f$. On the unit <sphere>, $\mathbf n=\mathbf r$ and $\mathbf F\cdot\mathbf n=f'(z)$. A spherical band between heights $z$ and $z+dz$ has <area> $2\pi\,dz$, so the curved <flux> is
$$
\Phi_{\mathrm{curved}}=2\pi\int_{-1}^{z_0}f'(z)\,dz
=2\pi\bigl(f(z_0)-f(-1)\bigr),
$$
by the <fundamental theorem of calculus>. On the horizontal top <disk>, $\mathbf F\cdot\mathbf n=z_0f'(z_0)$ and the <area> is $\pi(1-z_0^2)$. Thus
$$
\boxed{\Phi(z_0)=2\pi\bigl(f(z_0)-f(-1)\bigr)
+\pi z_0(1-z_0^2)f'(z_0)}.
$$
As usual for these <surface integrals>, the first <derivative> is assumed sufficiently integrable for the displayed <fundamental theorem of calculus> evaluation; a continuous first <derivative> suffices. Mere pointwise <differentiability> without such an integrability assumption need not make an ordinary <flux integral> exist.
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