In the following proof, I believe I have only showed that the $2$-form given equals the orientation form $\omega$ on the intersection $U_x\cap U_y\cap U_z$. I'm not sure if I can do the same `trick' involving $\frac{dy\wedge dz}{x}, \frac{dz\wedge dx}{y}$, and $\frac{dx\wedge dy}{z}$ in just the open sets $U_x, U_y$, and $U_z$ (no triple intersection) since these expressions are not defined away from where the denominator of each is nonzero. I am wondering whether I have written enough to show that the 2-form given really equals $\omega$ in $U_x, U_y$, or $U_z$ (away from the triple intersection) since the vanishing of $x,y,$ or $z$ are each a codimension 1 condition.
Exercise 23.5: Show that on $S^2\subset \mathbb{R}^3$, we have that the 2-form $xdy\wedge dz + y dz\wedge dx + z dx\wedge dy$ equals the orientation form
$$\omega = \begin{cases} \frac{dx \wedge dy}{z} & \text{ on }z\neq 0,\\ \frac{dy\wedge dz}{x} & \text{ on }x \neq 0,\\ \frac{dz\wedge dx}{y} & \text{ on }y\neq 0. \end{cases} $$
Proof $S^2\subset \mathbb{R}^3$ is defined as $S^2 := f^{-1}(0)$ where $f(x,y,z):= x^2+y^2+z^2-1$. Then $f=0$ implies $df = 2xdx+2ydy+2zdz=0$, so that \begin{align*} dx &= \frac{-ydy-zdz}{x} &\text{ on } x\neq 0, \\ dy &= \frac{-xdx-zdz}{y} & \text{ on }y\neq 0, &\text{ and }\\ dz &= \frac{-xdx-ydy}{z} & \text{ on }z\neq 0. \end{align*}
From these relations, we have on $U_x$ that $dx \wedge dy= \frac{-z}{x}dz\wedge dy$, and $dx \wedge dz = \frac{-y}{x}dy\wedge dz$.
On $U_y$, $dy\wedge dx = \frac{-z}{y}dz\wedge dx$ and $dy\wedge dz = \frac{-x}{y}dx\wedge dz$.
On $U_z$, $dz\wedge dx = \frac{-y}{z}dy\wedge dx$ and $dz\wedge dy = \frac{-x}{z}dx\wedge dy$.
From $\frac{dx\wedge dy}{f_z} = \frac{dy\wedge dz}{f_x} = \frac{dz\wedge dx}{f_y}$ on $U_x \cap U_y\cap U_z$ [construction of non-vanishing n-form on pre-image of regular value where $f_x, f_y$ and $f_z$ are partial derivatives] we have
\begin{align*} \frac{dy\wedge dz}{2x} = \frac{dz\wedge dx}{2y} = \frac{dx\wedge dy}{2z}. \end{align*}
Multiplying through by 2 gives
\begin{equation}\begin{aligned} \frac{dy\wedge dz}{x} = \frac{dz\wedge dx}{y} = \frac{dx\wedge dy}{z}. \end{aligned}\end{equation}
On $U_x\cap U_y\cap U_z$, we have
$$(xdy\wedge dz + ydz\wedge dx + zdx\wedge dy)|_{U_x} = x^2 (\frac{dy\wedge dz}{x}) + y^2(\frac{dz\wedge dx}{y}) + z^2 (\frac{dx\wedge dy}{z}) \\ = (x^2+y^2+z^2) ( \frac{dy\wedge dz}{x}) \hspace{2cm}\text{ by above equation}\\ = \frac{dy\wedge dz}{x} \text{ since } x^2+y^2+z^2=1.$$
This shows that $\omega = xdy\wedge dz + ydz\wedge dx + zdx\wedge dy$ on $U_x\cap U_y \cap U_z$, and since the substitutions in Equation (1) hold on any overlaps where the expressions are well-defined, we are free to make the same substitution in $U_x$, $U_y$ and $U_z$. End Proof