It's a bit hard to exactly construct a stress-energy tensor similar to a wormhole in normal space since part of the assumption is that the topology isn't simply connected, but consider the following scenario :
Take a thin-shell stress-energy tensor such that
$$T_{\mu\nu} = \delta(r - a) S_{\mu\nu}$$
with $S_{\mu\nu}$ the Lanczos surface energy tensor, where the Lanczos tensor is similar to a thin-shell wormhole. For a static spherical wormhole, that would be
\begin{eqnarray}
S_{tt} &=& 0\\
S_{rr} &=& - \frac{2}{a}\\
S_{\theta\theta} = S_{\varphi\varphi} &=& - \frac{1}{a}\\
\end{eqnarray}
If we did this by the usual cut and paste method (cutting a ball out of spacetime before putting it back in, making no change to the space), the Lanczos tensor would be zero due to the normal vectors being the same (there's no discontinuity in the derivatives). But we're imposing the stress-energy tensor by hand here. This is a static spherically symmetric spacetime, for which we can use the usual metric
$$ds^2 = -f(r) dt^2 + h(r) dr^2 + r^2 d\Omega^2$$
with the usual Ricci tensor results :
\begin{eqnarray}
R_{tt} &=& \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{f'}{rhf}\\
R_{rr} &=& - \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{h'}{rh^2}\\
R_{\theta\theta} = R_{\varphi\varphi} &=& -\frac{f'}{2rhf} + \frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h})
\end{eqnarray}
Using $R_{\mu\nu} = T_{\mu\nu} - \frac 12 T$ (this will be less verbose), we get that $T = -\delta(r - a) [2(ah)^{-1} + 2 (ar^2)^{-1}]$, and then
\begin{eqnarray}
\frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{f'}{rhf} &=& \delta(r - a) \frac{1}{a} (\frac{1}{h} + \frac{1}{r^2}) \\
- \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{h'}{rh^2} &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]\\
-\frac{f'}{2rhf} + \frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h}) &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]
\end{eqnarray}
This is fairly involved and I'm not gonna solve such a system, so let's make one simplifying assumption : just as for the Ellis wormhole, we'll assume $f = 1$, which simplifies things to
\begin{eqnarray}
0 &=& \delta(r - a) \frac{1}{a} (\frac{1}{h} + \frac{1}{r^2}) \\
\frac{h'}{rh^2} &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]\\
\frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h}) &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]
\end{eqnarray}
The only solution for the first line would be $h = - r^2$, but then this would not be a metric of the proper signature. I don't think there is a solution here (or if there is, it will have to involve a fine choice of the redshift function), which I believe stems from the following problem :
From the Raychaudhuri equation, we know that in a spacetime where the null energy condition is violated, there is a divergence of geodesic congruences. This is an important property of wormholes : in the optical approximation, a wormhole is just a divergent lense, taking convergent geodesic congruence and turning them into divergent ones. This is fine if the other side of the wormhole is actually anoter copy of the spacetime, but if this is leading to inside flat space, this might be a problem (once crossing the wormhole mouth, the area should "grow", not shrink as it would do here).
A better example, and keeping in line with the bag of gold spacetime, is to consider a thin-shell wormholes that still has trivial topology. Take the two manifolds $\mathbb R^3$ and $\mathbb S^3$. By the Gauss Bonnet theorem, a sphere must have a part in which it has positive curvature (hence focusing geodesics). Then perform the cut and paste operation so that we have the spacetime
$$\mathcal M = \mathbb R \times (\mathbb R^3 \# S^3)$$
Through some topological magic, this is actually just $\mathbb R^4$. The thin-shell approximation is easily done here, and it will give you the proper behaviour : geodesics converge onto the mouth, diverge upon crossing the mouth, then go around the inside of the sphere for a bit before possibly getting out.
From there, it's possible to take various other variants, such as smoothing out the mouth to make it more realistic (which will indeed give you a bag of gold spacetime), as well as a time dependancy to obtain this spacetime from flat Minkowski space.
It's true that traversable wormholes would theoretically allow for travel into the past according to general relativity (though this may be ruled out by quantum effects in whatever theory of quantum gravity eventually replaces general relativity, see the chronology protection conjecture). But the only way this would happen is if one mouth of the wormhole experiences time dilation relative to the other, either because it's taken on a journey at relativistic speed relative to the other mouth, or because it's moved closer to a source of gravity where it experiences gravitational time dilation. If the two mouths have clocks moving alongside them which were initially synchronized, then if some time dilation has accumulated, when the two mouths are brought back together an outside observer looking at them side-by-side will see the clocks showing different times. But, as explained by physicist Kip Thorne (who discovered traversable wormholes as a theoretical possibility in general relativity) in his book Black Holes and Time Warps, time threads differently through the wormhole, in such a way that if you look through one mouth at the clock which is alongside the other mouth, and compare it with the clock alongside the mouth you're looking through, then the clocks will still be synchronized (assuming the time for light to travel through the wormhole from the clock to your eyes is negligible). This means for example that if a side-by-side comparison by an outside observer (one who was not looking through either wormhole) showed mouth #1's clock to read 2015 and mouth #2's to read 2010, then if you jumped through mouth #2 you would exit mouth #1 when its clock also read 2010, not 2015. In some cases this would make it so that when you jump through a wormhole, you could end up in a region of spacetime where it would be possible to send a signal to your own younger self, which would be received at an earlier time then it was sent according to a clock you carried along with you (i.e. in terms of your own proper time).
But you don't mention anything about time dilation--if one mouth of the wormhole is moved slowly from A to B, so that no time difference accumulates, then there will be no time travel in your scenario. One thing to keep in mind is that in relativity, simultaneity is not generally defined in terms of when you see the light from events--for example, in an inertial frame in special relativity, if I see the light from an explosion 10 light-years away in the year 2020 (in the space and time coordinates of that frame), I subtract out the light travel time and say the event actually took place in 2010 in my frame. And wormhole spacetimes can be asymptotically flat, meaning to a good approximation you can treat them as localized distortions of spacetime moving around in an otherwise flat spacetime, so you can still set up something very close to an inertial frame in the region outside the wormholes. In this case, we might say that in a frame where A and B are at rest, the event of the clocks at A and B each reading 12:00:00 AM on Jan 1. 2020 are simultaneous. Then if I jump into mouth B at that moment, and exit from mouth A 4 seconds later according to the time coordinates of this frame, we can predict that I will see the clock at A reading 12:00:04 AM, Jan. 1 2020. If I immediately look back through the wormhole I will see the clock at B reading the same time of 12:00:04 AM, Jan. 1 2020, but if I immediately look at B through normal space using a telescope, due to the 10-light-year distance I will see the clock at B reading 12:00:04 AM, Jan 1. 2010. Assuming I had been living near B until I jumped through that mouth in 2020, looking through my telescope will also allow me to see myself when I was ten years younger.
But the main thing to realize is that there's no way I can actually send a signal to reach my younger self in the past in this scenario, as was possible in the other scenario where time dilation had created a time difference between clocks next to each mouth. If I send it through the wormhole, then (assuming a negligible travel time for a light signal) it exits mouth B 12:00:04 AM Jan. 1 2020, whereas I had jumped through mouth B 4 seconds earlier at 12:00:00 AM Jan. 1 2020. Meanwhile, if I send a light signal through normal space at 12:00:04 AM Jan 1 2020, in my frame it takes 10 years to get there, and another 10 years for the light from the event of someone receiving it at B to get back to me. This means I won't see anyone at B receive it (at least not if I am looking through normal space) until my clock reads 12:00:04 AM Jan 1 2040, and I will see the clock at B reading 12:00:04 AM Jan. 1 2030 when someone next to it receives my signal. So regardless of whether I send a signal through normal space or back through the wormhole, in this scenario the signal will arrive at B when the clock there shows a later time than the moment I jumped through the wormhole to travel from B to A.
Best Answer
Wormholes don't just exist by themselves, they have to be created by a form of matter called exotic matter (which almost certainly doesn't exist, but let's gloss over this). To construct a wormhole you need to gather up some exotic matter and arrange it in a particular configuration.
So if the exotic matter is moving the wormhole will move along with it. However you may find that a considerable force is required to move the matter at the two ends of the wormhole because as they move energy will have to be put into the spacetime warping between them.
Re the second question: a wormhole is not like a Star Trek transporter that instantly moves you from one place in space to another. If you pass through a wormhole you would be floating freely$^1$ just like the astronauts in free fall in the International Space station. Falling through the spacetime in a wormhole is no different to falling through the spacetime around the Earth, except that the spacetime in a wormhole has a rather special shape.
Generally speaking, falling freely through spacetime does change your velocity (as measured by someone who is not falling freely). For example, when Felix Baumgartner jumped out of his capsule his speed relative to the ground changed from zero to Mach 1.25. In that case it was the spacetime curvature caused by the mass of the earth that accelerated him. In the case of the wormhole it would be the spacetime curvature caused by whatever exotic matter you used to construct the wormhole.
So when you emerge from the far end of the wormhole your speed will almost certainly be different to the speed you had when you jumped in. Exactly what your speed will be depends on the exact geometry. The speed will make absolutely no difference to you unless there is something in your way, so putting the far end in a closed room would be a bad (and very messy :-) mistake. In any of the more realistic science fiction stories that use wormholes (and remember, this is science fiction we're talking about) the ends of the wormholes are floating in space so anything emerging from them has lots of room to slow down.
$^1$ though there maybe tidal forces