If we want to be pedantic about it (and if we don't want to say that the question is trivial), the statement of "velocity reciprocity" does not make sense on its face in a fully general setting: Mathematically, positions are points in the space(time) manifold, and velocities are tangent vectors attached to these points; saying that $v_1 = -v_2$ for two velocity vectors only makes sense when they're attached to the same point, otherwise these are simply incomparable objects. When you attach reference frames to two moving different objects, unless the objects are currently colliding, we therefore first need to establish what the statement that their velocities are "equal" actually means:
What allows us to compare tangent vectors at different points is parallel transport. In general, this transport depends on a path, but in flat space, it does not, and this explains why all this stuff about tangent vectors usually doesn't play a role in introductory physics course: In flat space, we can just identify all the tangent spaces with each other, since there's a unique flat parallel transport operator between them.
In curved space, the question becomes ill-defined as there is no unique definition of relative velocities in that case since parallel transport varies with the path we might choose (see e.g. this excellent answer by A.V.S).
So, assuming we're in flat space, there isn't really any problem: The simplest way to implement parallel transport there is just to draw a grid of straight lines (i.e. geodesics) on the floor. Then both your frames have a coordinate system given by the tangent vectors of these geodesics (and additionally the grid makes measuring velocity easy with a stopwatch and some geometry), and you can tell the people in the frames to measure velocity and you'll get $v_1 = -v_2$.