I know the way that basically everywhere says to do it, but I'm having trouble putting it into equations that make sense for logarithmic branch points.
Everywhere says to trace a path around the branch points, with the angles $\theta_1$ and $\theta_2$ corresponding to the angles between each branch point and any given point on the path. As a circle is traced around the first branch point, $\theta_1$ gets incremented by $2\pi$, then the path is followed along the cut to the next branch point with the angles remaining the same. Then a circle is traced around the second branch point, with $\theta_2$ getting incremented by $2\pi$. Then you follow the cut back to the initial point, again with no change in angle.
If the function is single valued, you should arrive back at the same value for $f(z)$ after tracing this path. This is easy to show for something like $((z-1)(z+1))^{1/2}$ :
$$ (r_1e^{i\theta_1} \cdot r_2e^{i\theta_2})^{1/2}=r_1r_2e^{i(\frac{\theta_1+\theta_2}{2})} \\ $$
$$ \begin{align*} (r_1e^{i(\theta_1+2\pi)} \cdot r_2e^{i(\theta_2+2\pi)})^{1/2}& =r_1r_2e^{i(\frac{\theta_1+\theta_2}{2}+2\pi)} \\ &=r_1r_2e^{i(\frac{\theta_1+\theta_2}{2})}e^{i(2\pi)} \\ &=r_1r_2e^{i(\frac{\theta_1+\theta_2}{2})}(1) \\ &=r_1r_2e^{i(\frac{\theta_1+\theta_2}{2})} \end{align*} $$
But throw in a logarithm with infinite sheets and it doesn't seem to add up, like: $\log((z-1)(z+1))$
$$ \log(r_1e^{i\theta_1} \cdot r_2e^{i\theta_2}) = \ln(r_1r_2) + i(\theta_1+\theta_2) $$
$$ \log(r_1e^{i(\theta_1+2\pi)} \cdot r_2e^{i(\theta_2+2\pi)}) = \ln(r_1r_2) + i(\theta_1+\theta_2 +4\pi) $$
It's no longer periodic so multiples of $2\pi$ no longer cancel out automatically.
If you set $0\leq\theta_1<2\pi$ and $-\pi\leq \theta_2 < \pi$ which i believe it is supposed to be with the additional branch point at infinity, seemingly the only way to stay single valued going from $\theta_1=0 $ to $\theta_1=2\pi$ and simultaneously from $\theta_2=-\pi $ to $\theta_2=\pi$ is for the trace to cross over the branch cut.
$$ \begin{align*} \log(r_1e^{i(0)} \cdot r_2e^{i(-\pi)})& = \ln(r_1r_2) + i(-\pi) \\ \log(r_1e^{i(0+2\pi)} \cdot r_2e^{i(-\pi+2\pi)}) &= \ln(r_1r_2) + i(3\pi) &\text{ (not the same value)}\\ \end{align*} $$
$$ \begin{align*} \\ \log(r_1e^{i(0)} \cdot r_2e^{i(\pi)})& = \ln(r_1r_2) + i(\pi) \\ \log(r_1e^{i(0+2\pi)} \cdot r_2e^{i(\pi-2\pi)}) &= \ln(r_1r_2) + i(\pi) &\text{ (same value)}\\ \end{align*} $$
Basically it seems like one of the branch points has to be traced around the wrong direction, from the maximum angle to the minimum angle, which corresponds to the trace crossing over the branch cut before going around the point. If you treat infinity as a single point on the Reimann sphere, tracing around the cut means both points should be gone around either clockwise or both counterclockwise, but that seems to result in extra $i\pi$'s in the value of the function.
What is going on here? I'm clearly missing something because I know this works, just every explanation I have seen is lacking some kind of information needed to describe the full idea of what they are trying to do here.
Since there seems to be some confusion, this is the method for testing a branch cut I was referring to earlier:
Maybe we can actually draw some pictures and clear up your confusion. $f(z) = \log(z^2-1)$ has three branch points where it is not analytic: $z=1, z=-1, z=\infty$.
In order to make $f$ single-valued, we need to restrict it to a simply-connected domain on the Riemann sphere. We can do that by connecting the three branch points by a simple curve, and discarding that curve from the domain of $f$. While any simple curve would work, it is almost always easiest to use lines:
In this example, $1$ is connected to $-1$, and $-1$ is further connected to $\infty$. The domain $U$ consisting of all the Riemann sphere outside this curve is simply-connected. We can choose a particular value for $f$ at some point not on $(-\infty, 1]$, say $f(\sqrt{e + 1}) = 1$. Then take the Taylor series of $f$ about this point, having the selected value as constant term, and arrive at a single-valued function within the radius of convergence. Then per the result reuns quoted, extend that function to the entirety of $U$, on which by Cauchy's theorem it will be single-valued, since $f'$ is analytic everywhere in $U$ and thus will integrate to $0$ around any closed loop.
Now, let's add your path description:
I hope you can see why that didn't work. The whole point of cut lines is that they are removed from the domain of the function. You cannot cross them. When you cross the cut, the value of the function jumps discontinuously, which you ignored.
If you want examine it properly, you have to also draw a circle around the third branch point, $\infty$. When projected into the plane, a small counter-clockwise circle about $\infty$ becomes a large clockwise circle about $0$, which encircles both of the other branch points.