Take a close look at your first diagram and consider what would be different if you had drawn the position of point $B$ a little bit lower on the page, but otherwise kept the directions of the vectors and the cones the same.
The answer is: everything would be the same except that $B$ would be
lower on the page and the angle of the line between $A$ and $B$ would be different.
If the information we have can be made to fit different possible angles
equally well, we have no way to tell exactly what the angle actually is.
The good news is, if you know precisely the directions of the boundaries of both cones, you have enough information to say precisely what the range of possible angles from $A$ to $B$ might be.
I'll assume the two known vectors are parallel, or if they are not, that we known the angle between them. If the vectors are parallel, we call that
direction "north"; if they are not parallel we choose one of them as the "north" direction and (much later) use the knowledge of the angle between them to figure out the part of the answer that is relative to the other vector.
For each vector or ray in the figure, I'll consider its "course angle" to be the angle clockwise from "north" to the given angle or ray.
I also assume that we now have enough information to determine the course angles of the two "edges" of each $15$-degree cone, and that we have computed those angles.
So the edges of the cone at $A$ give a range of possible course angles
of the vector $v_{AB}$ that points in the direction from $A$ to $B,$
and the edges of the cone at $B$ give a range of possible course angles
of the vector $v_{BA}$ that points in the direction from $B$ to $A.$
But we also know that the vector $v_{BA}$ is exactly $180$ degrees opposite from vector $v_{AB}.$ So the information about the range of course angles for vector $v_{BA}$ is also information about the range of possible course angles for vector $v_{AB}.$
For example, suppose we find that the edges of the cone at $A$ are the course angles $60$ degrees and $75$ degrees.
That is, the course angle $\theta_{AB}$ from $A$ to $B$ must be somewhere in the range $60^\circ \leq \theta_{AB} \leq 75^\circ.$
And suppose the edges of the cone at $B$ are the course angles
$235$ degrees and $250$ degrees.
Then the course angle $\theta_{BA}$ from $B$ to $A$ must be in
the range $235^\circ \leq \theta_{BA} \leq 250^\circ.$
But since the two courses are $180$ degrees opposite,
the cone at $B$ tells us that
$235^\circ \leq \theta_{AB} +180 \leq 250^\circ,$
or equivalently,
$55^\circ \leq \theta_{AB} \leq 70^\circ.$
So we can make a better estimate of the course from $A$ to $B$ by using
both cones than by using one:
$60^\circ \leq \theta_{AB} \leq 70^\circ.$
If we choose the middle of that range, $65$ degrees, we minimize
the possible amount of the error in the course angle.
If the two cones are always aligned so that their edges are parallel to the other cone's edges, however (as it appears may be the case in the diagrams in the question), then we don't get any new information by considering both cones that we didn't have from looking at just one cone.
As for how much new information you would need to make the measurement exact, the simplest piece of information would be the actual course angle from $A$ to $B.$ Other possibilities are to have coordinates of the locations of both points, which would let you calculate a course angle with trigonometry--but that's a lot of information about the points (you now know exactly where each of them is!) and it wouldn't use the cones at all.
We could also ask for two cones such that the "left" edge of one cone was parallel to (or $180$ degrees opposite from) the "right" edge of the other cone; but that's just a very thinly disguised way of asking to be given the exact course angle (it's the same as the angle of one of the edges we asked to be aligned).
