Evaluating the integral $\int_{-1}^{1} \frac{\sqrt{1-x^{2}}}{1+x^{2}} \, dx$ using a dumbbell-shaped contour

You really don't want to break the square root up; it's not about separate branch cuts, but about a single branch cut for the square root of the quadratic. Presumably, you want the branch of $g(z)=\sqrt{1-z^2}$ with $g(i)=+\sqrt2$. When you then compute the residue of $f$ at $\infty$, you have to interpret the residue at $0$ of $$\sqrt{1-\big(\frac1u\big)^2} = \frac{\sqrt{u^2-1}}{\sqrt{u^2}},$$ and one has to check that the ambiguities in the two square roots cancel out.

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Instead of confused about what $\arg(−1)$ is, go back to original integral and observe for given choice of argument, $\arg\sqrt{1-x^2} = \frac{\pi}{2}$ for $x > 1$ on real axis. As a result, $\displaystyle\;\frac{\sqrt{1-z^2}}{1+z^2} \sim \frac{i}{z}$ for large $z$.

Deform the clockwise dumbbell contour "continuously" to a clockwise circular contour at infinity. In the middle of process, one pickup two extra counterclockwise circular contours at poles ($\pm i$) of the integrand.

The contribution from the clockwsie circular contour at infinity is controlled by the large $z$ behavior of the integrand. It equals to $$(-2\pi i)i = 2\pi$$

Since $\arg\sqrt{1-z^2}$ is negative on positive imaginary axis and positive on negative imaginary axis, contribution from the two counterclockwise circular contours around the poles is

$$(2\pi i)\left(\frac{-\sqrt{2}}{2i} + \frac{\sqrt{2}}{-2i}\right) = -2\pi\sqrt{2}$$ This leads to

$$-2I = -2\pi\sqrt{2} + 2\pi \quad\implies\quad I = \pi(\sqrt{2}-1)$$