I'm not sure what you mean by complex-angular measure, but there is indeed a geometric interpretation.

Suppose $z$ and $w$ have homogeneous coordinates $[z_0,z_1]$ and $[w_0,w_1]$, respectively. Then
$$
U(z,w) = 1 + \frac{z_1}{z_0} \frac{\overline{w}_1}{\overline{w}_0},
$$
and hence
$$
\widetilde{U}(z,w) = \frac{\overline{z}_0}{|z_0|} \frac{w_0}{|w_0|}
\frac{z_0 \overline{w}_0 + z_1 \overline{w}_1}{\sqrt{(|z_0|^2+|z_1|^2)(|w_0|^2+|w_1|^2)}}.
$$
So the question is really asking what the complex phase of $\overline{z}_0 w_0 (z_0 \overline{w}_0 + z_1 \overline{w}_1)$ means.

To put this in a more general setting, let's look at $\mathbb{C}\mathbb{P}^{n-1}$. We'll represent points by unit vectors in $\mathbb{C}^n$, and we'll use the Hermitian form $\langle \cdot, \cdot \rangle$ on $\mathbb{C}^n$. As you pointed out, the absolute value of the inner product $\langle x,y \rangle$ measures the distance between $x$ and $y$ in $\mathbb{C} \mathbb{P}^{n-1}$. The complex phase of $\langle x,y \rangle$ is not well-defined, because we can rotate $x$ and $y$ by unit complex numbers without changing the points in projective space. However, the phase change along a cycle of points is invariant: in the expression
$$
\langle x,y \rangle \langle y,z \rangle \langle z,x \rangle,
$$
phase rotations for $x$, $y$, or $z$ won't change the answer (because they occur once on the linear side of the Hermitian form and once on the conjugate-linear side).

I'm not sure what to call this quantity. I think it's more or less the Pancharatnam phase from physics, but that's a little far afield from what I know. In any case, it gives an invariant for $k$-tuples of points under the action of the unitary group $U(n)$. For two points, you just get $\langle x,y \rangle \langle y,x \rangle = |\langle x,y \rangle|^2$, so it gives the Fubini-Study metric again. For more than three points, you get a perfectly good invariant, but it's not as essential because it can generically be reduced to $2$- and $3$-point invariants. For example,
$$
\langle w,x \rangle \langle x,y \rangle
\langle y,z \rangle \langle z,w \rangle =
\frac{\langle w,x \rangle \langle x,y \rangle
\langle y,w \rangle
\cdot \langle y,z \rangle \langle z,w \rangle
\langle w,y \rangle
}{\langle y,w \rangle \langle w,y \rangle},
$$
as long as $w$ and $y$ aren't orthogonal. However, when $\langle w,y \rangle = \langle x,z \rangle = 0$, the $4$-point invariant is not determined by lower-order invariants (the $3$-point invariants all vanish, and the $2$-point invariants don't determine the complex phases at all).

The unitary group $U(n)$ maps one $k$-tuple of points in $\mathbb{C} \mathbb{P}^{n-1}$ to another if and only if all the invariants of corresponding sub-tuples agree, and as long as there are no orthogonal pairs of points, it suffices to get agreement for pairs and triples. So these invariants completely classify the orbits of $U(n)$ acting on finite subsets of $\mathbb{C}\mathbb{P}^{n-1}$.

The phase of $\widetilde{U}(z,w)$ in the original example is a special case of this invariant. It's the $3$-point invariant of $[z_0,z_1]$, $[w_0,w_1]$, and $[1,0]$ (assuming the Hermitian form is conjugate-linear in the second variable). Specifically, that gives
$$
(z_0 \overline{w}_0 + z_1 \overline{w}_1) w_0 \overline{z}_0,
$$
as desired. The appearance of $[1,0]$ here is a little arbitrary, but that's just a consequence of how the original problem was set up.

I've cheated a bit in one respect: the question asked about $\mathbb{C}\mathbb{P}^1$, and on $\mathbb{C} \mathbb{P}^1$ the higher-order invariants are not so exciting, because the orbits of $U(2)$ on arbitrary $k$-tuples of points are simply determined by the pairwise distances between the points. (It's essentially the same as $SO(3)$ acting on $S^2$, in which case this is a familiar fact.) So if we restrict attention to $\mathbb{C}\mathbb{P}^1$, then it's still true that the answer is a $3$-point invariant, but it's not really giving any new information beyond the Fubini-Study metric. However, in higher dimensions it does.