This first section is not part of an answer.
Perhaps it would be useful to see a Riemann surface of $f$. (These plots are generated by Jason B.'s code from this answer at Mathematica.SE, which is based on Michael Trott's RiemannSurfacePlot3D.) First, on a disk centered at $0$ of radius $3/2$ with the real axis running right-left and the imaginary axis increasing into the screen.
The branch points at $-1$, $0$, and $1$ are visible from left to right. Note that this is a $3$-sheeted (branched) covering of $\Bbb{C}$.
Now a Riemann surface for $f(1/z)$. The real axis runs from slightly above center-left to slightly below center-right. The imaginary axis runs into the screen from the lower-left to the upper-right. The branch points at $-1$ and $1$ are visible as the left and right dimples, respectively, in the top layer.
The $3$-sheetedness continues to $\infty$ and is expressed by trifolium cross-sections in the plot.
Perhaps it would help to see a Riemann surface for $(z^3)^{1/3}$, which is not as simple as one might expect.
and $((1/z)^3)^{1/3}$ to see the neighborhood of $\infty$.
Now maybe some response to the Question...
I am not familiar with Osborne's text, but I am not happy with the quoted discussion of branch cuts. A branch cut can be an arbitrary simple (i.e., non-self-intersecting) curve, with each endpoint on a different branch point.
A line segment is a line segment -- a finite segment of a line. This particular characterization of a branch cut is inadequate in many respects. One is that it cannot represent any of the choices of branch cut for the logarithm, which are necessarily infinitely long since they must reach from the branch point at $0$ to the branch point at $\infty$.
A branch point cannot be met by a branch cut between two other branch points. At such a hypothetical intersection, the values of all the branches are the same (since we are at a branch point), but the values of the function approaching the branch point from either side of the branch cut disagree, which is a contradiction.
We should perhaps distinguish between branches and choices of values of the function. For a multivalued function, at each point, we may select one of its values to associate with that point. An easy example: for each $x \in \Bbb{R}_\geq 0$ assign an arbitrary choice of $\sqrt{x}$ or $-\sqrt{x}$. Note that we have many such collections of choices. Only two of these collections are branches of the square root function. Branches have the property that one can analytically continue along them. This means that in any small enough neighborhood of a branch point, no branch cut other than the branch cut terminating at that branch point may be present. This is so we can analytically continue the branch around the point to get a branch, not just a bunch of choices of values.
You rightly want $\Bbb{C} \smallsetminus (\{\text{branch cuts}\} \cup \{\text{branch points}\})$ to be connected (in fact, path connected). This also means that branch cuts cannot intersect transversely: If two cuts intersect, there is at least one region (homotopic to a disk) bounded by the branch cuts and separated from the rest of the plane, so there is no way to analytically continue from the interior of the disk to the rest of $\Bbb{C}$ or vice versa. Allowing transversely intersecting branch cuts more closely corresponds to a choice of value than to a choice of branch. The same thing happens if you allow two (nonintersecting) cuts between two branch points: there is now a disk bounded by the two cuts and two points which is disconnected from the rest of $\Bbb{C}$. I suppose one could have two branch cuts meet non-transversely, but a small homotopy of one would eliminate the intersection.
Returning to your $f$. This means you must have a branch cut between two of your branch points and another branch cut from the third branch point to the branch point at infinity. That there is a branch point at infinity can be shown by $\lim_{z \rightarrow 0} \frac{1}{f(1/z)} = 0$ (That is, $f$ is single valued at $z = \infty$ on the Riemann sphere.), just as one does for the logarithm. That it is of order $3$ can be seen by counting the number of sheets of the Riemann surface in neighborhoods of $z = \infty$.
As more evidence, the intersections of the folium suggest something interesting happens for arguments near $\pm \pi/3$ and $\pm \pi$. (The curves of interest are asymptotic to the rays of argument $\pm \pi/3$, and finding them is a bit of a hassle, so we look at what happens around $\pm \pi$.) Transverse to the negative real axis, choosing a branch with cuts along $(-\infty, -1)$ and $(0,1)$, so that we may continue from $-1/2$ to the points in the limits below, \begin{align*}
\lim_{\theta \rightarrow -\pi^+} f \left( 10 \mathrm{e}^{\mathrm{i} \theta} \right) &= -(-3)^{2/3}110^{1/3} = 4.9832{\dots} + 8.6312{\dots} \mathrm{i} \\
\lim_{\theta \rightarrow \pi^-} f \left( 10 \mathrm{e}^{\mathrm{i} \theta} \right) &= 3^{2/3}(-110)^{1/3} = 4.9832{\dots} - 8.6312{\dots} \mathrm{i}
\end{align*}
So when the cut switches sheets along the negative real axis, the argument decreases by $2\pi/3$ as we cross it in the positive argument direction. So, at the very least, $f(z)$ does not return to its value after orbiting along the circle of radius $10$ from argument $-\pi$ to argument $\pi$, crossing no branch cuts while doing so. It takes three orbits for the argument of the value of $f$ to decrease by $2\pi$, replicating the starting value.
