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.
First let me say that wormholes, event horizons and many other possible effects associated with collapsed stars are currently purely theoretical. We think these effects and other associated events may result from the collapse of a star with a mass in excess of the Chandrasekhar Limit
We are fairly sure we are on the right track as to what what those effects are because we use GR to predict them, and it is a theory that, so far at least, has passed every test we have thrown at it.
Unfortunately we have as yet no direct proof of these exotic effects, and experts in General Relativity (and I am NOT one of them) may sometimes disagree about the detailed physics of black holes.
Any ideas about wormholes allowing you to move to another universe, or different places within our universe, are currently predictions using the math of GR , some of which may well be correct but none of which we have any proof of, as of today.
General Relativity has been confirmed very well in "normal" situations, such as predicting the changes in orbit of Mercury, and we use it's predicted effects in GPS systems, but when it comes to extreme events, such as black holes, we are predicting what "might" happen, more than currently observing what is actually happening.
If a wormhole exists in the gravity well of a star is the same as saying, if a blackhole singularity "appears" near a star.
Well, the first thing that will probably happen is that the black hole will pull the normal star towards it, possibly creating what is called an accretion disc of material from the other star.
We are pretty sure we have evidence that this actually happens. In the worse case (for the normal star), it will be pulled apart by the intense gravity of the black hole.
Anything located at point B, will feel the effects of the black hole, and a wormhole (if such a thing actually exists) will affect anything at point B, just like a normal star would, there may not be anything much exotic about it, just ordinary gravity.
So , in that sense, will the gravity effects travel through the wormhole, is the same as saying point B feels the effects of gravity.
The "wormhole", that is the point of intense gravity, will act in exactly the same way as the Sun pulls the Earth towards it.
Best Answer
Disclaimer: I'm not a GR expert, but this is how this question has been explained to me by other physicists before. If I got something wrong, please correct me.
The traveler does indeed not have to exert as much work to leave the gravity well via the wormhole compared to the normal route. They are not repelled from mouth A nor attracted to mouth B by any effect having to do with the gravity of the planet.
Conservation laws are preserved, however, by interaction with the wormhole mouths themselves. When the traveler enters mouth A and leaves mouth B, no work is required to raise their mass because mouth A appears to gain equal mass to the traveler, and mouth B loses it. As far as conservation laws are concerned, it's as if the traveler crashed into and merged with an asteroid in low orbit (mouth A), and then an identical copy of the traveler got assembled out of the mass of another asteroid (mouth B) and ejected in high orbit.
So, if you try to generate infinite energy by throwing something through the wormhole and then running a generator off it as it falls back down, your plans will be foiled by mouth A becoming steadily more massive while mouth B becomes steadily less massive, until mouth A collapses into a black hole.