Let us change notation somewhat: Let $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ be points in $\mathbb R^n$ such that $|x|=|y|=1$ and $|x-y|=d\in(0,1)$, where $|\cdot|$ is the Euclidean norm. Let $v$ be a random vector uniformly distributed on the unit sphere $S^{n-1}$ in $\mathbb R^n$. The probability in question is
\begin{equation}
p:=P(x\cdot v<0<y\cdot v)+P(y\cdot v<0<x\cdot v)=2P(x\cdot v<0<y\cdot v),
\end{equation}
where $\cdot$ is the dot product.

The key note is that the random vector $v$ equals in distribution the random vector $(Z_1,\dots,Z_n)/\sqrt{\sum_1^n Z_i^2}$, where $Z_1,\dots,Z_n$ are independent standard normal random variables (r.v.'s). Hence,
\begin{equation}
p=2P(X<0<Y),
\end{equation}
where $X:=\sum_1^n x_i Z_i$ and $Y:=\sum_1^n y_i Z_i$. The r.v.'s $X$ and $Y$ are jointly normal with zero means, unit variances, and correlation
\begin{equation}
r=EXY=x\cdot y=\tfrac12\,(|x|^2+|y|^2-|x-y|^2)=1-d^2/2.
\end{equation}
So, the pair $(X,Y)$ equals $(X,rX+\sqrt{1-r^2}\,Z)$ in distribution, where $Z$ is a standard normal r.v. independent of the standard normal r.v. $X$.

So,
\begin{equation}
p=2P(X<0<rX+\sqrt{1-r^2}\,Z)=2P(X<0,Z>-kX)=2P\big((X,Z)\in A\big),
\end{equation}
where
\begin{equation}
k:=r/\sqrt{1-r^2}
\end{equation}
and $A$ is the angle between the rays $\{(x,0)\colon x\le0\}$ and $\{(x,-kx)\colon x\le0\}$, emanating from the origin.
Since the distribution of the random vector $(X,Z)$ in $\mathbb R^2$ is rotation invariant, we conclude that the probability in question is
\begin{equation}
p=\frac\theta{\pi},
\end{equation}
where
\begin{equation}
\theta:=\text{arccot}\,k=\arccos r=\arccos(1-d^2/2)
\end{equation}
is the measure of the angle $A$.

In particular, it follows that $p=2d/\pi+O(d^3)\sim 2d/\pi$ as $d\downarrow0$, which agrees with the intuition.