What is distance and velocity regarding to wormholes? When we talk about Physics (Einstein's) laws we use notions of distance and velocity. Take for example objects A and B. In A's base at specific moment of time there's definite distance to B and definite B's velocity.
But if there's a short wormhole staring near A and ending near B there's no more definite distance. There're actually two distances from A to B - e.g. long and short.
If the B's end of wormhole moves at B's speed (in A's base) then there're also two B's velocities - one normal and one equal to zero measured via wormhole.
Doesn't that mean that notions of distance and velocity are incorrect in the space with wormholes, i.e. multiconnected space?
== UPDATE ===========================================
It seems that I should reformulate my question in a more strict way.
There're two relativity point objects A and B. There's also a wormhole W.
First wormhole mouth Wa is near A and is connected to A. Second mouth Wb is near B and it is moving together with B at B's speed in A's base. But the wormhole length (the distance between Wa and Wb measured through wormhole) is constant.
Now in A's base we have some relativity formulas for B's energy, time shift, etc. In these formulas we put Vb - B's velocity in A's base.
The question is: how do we measure Vb to use in the formulas?
There're two ways:


*

*We measure Vb in ordinary way out of wormhole. In this case Vb have some large value.

*We measure Vb through the wormhole. In this case Vb is equal to 0.
Which value should we use in the formulas?
 A: You don't need a wormhole for there to be different distances between two points, because gravitational lensing does that already.
In its most symmetric form the light flow round a gravitational lens looks something like:

The light from the object can follow one of two paths $a$ or $b$ shown by the solid lines to reach the observer, and as a result the observer sees two images. Actually the observer sees a ring, because the setup is axially symmetric about the line between the observer and object. This is the famous Einstein ring. Note that the paths $a$ and $b$ look curved to us because we're looking at a curved spacetime. As far as the light  rays are concerned the paths are straight. So we have two different straight lines from the object to the observer.
In this most symmetric setup the length of all the paths $a$, $b$ and the ones out of the plane of the diagram are the same. However break the symmetry slightly and now the paths $a$ and $b$ have different lengths. Let me emphasise this point. Light travels along straight lines - null geodesics - and yet there are multiple different paths to get from the object to the observer, and they can all have different lengths.
So we have exactly the same situation as you describe for your wormhole, though obviously the difference is far larger with a wormhole. My point is that this is routine in general relativity and not regarded as anything special. The notion that there is a unique distance between two points simply doesn't apply in GR - it is part of the baggage that you have to abandon.
Response to comment:
You assume that two objects taking the long way round and going through the wormhole would end up with different velocities. The trouble is that an object passing through the wormhole would experience acceleration because the spacetime in a wormhole is not flat so the transit through the wormhole would change it's velocity. This shouldn't be surprising as the wormhole has to be held open by enormous masses of exotic matter.
But to calculate how an object moves in the wormhole you would need to write down a metric, calculate its geodesics and likewise for the long way round. However we don't have any analytic metrics describing traversible wormholes so we could only do this calculation numerically.
However, provided the geometry is time independant we can be sure that energy is conserved - Noether's theorem tells us this. That means it's impossible to go round the loop and end up with more energy than you started with, and therefore that whichever way round you went you would end up with the same kinetic energy and therefore velocity.
