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.
This seems to be called the eternal-time-machine spacetime, and I believe the original paper was Morris 1988, which is available online and not paywalled. On p. 1447, they claim:
...at late times by traversing the wormhole from right mouth to left, one can travel backward in time (i.e., one can traverse a closed timelike curve)...
The question says:
I've read conflicting info -- some say that entering y will send the traveller back in time. Others say entering x will send the traveller back in time.
I'm not aware that this aspect of the idea is controversial at all. In your notation, the paper is saying that entering y and emerging at x sends you back in time. If there's some source that says it's the other way around, please tell us what the source is.
Here's a diagram showing what I understand them to be saying.
The wormhole is created before the diagram begins. The left-hand black line is the world-line of one mouth (x), and the other black line is the world-line of the other mouth (y). An observer's world-line ABCD is shown in red. B is the point where the observer enters y. C is the point where he exits x. The observer's world-line contains a closed, timelike curve. In the paper, they label the wormhole's mouths with time coordinates, which I think are the readings on a clock that's inside the wormhole. Since the wormhole can be assumed to be internally short, one can synchronize these clocks without any of the usual ambiguities -- even relativistically, one can absolutely synchronize two clocks that are at the same location. So B and C are simultaneous according to a synchronization process that's carried out inside the wormhole.
Morris, Thorne, and Yurtsever, "Wormholes, time machines, and the weak energy condition," Phys Rev Lett 61 (1988) 1446; http://authors.library.caltech.edu/9262/
Best Answer
I found a post here on physicsforums.com which has some useful links in the post by "pervect". One is to this article by physicist John Cramer, saying that each time a mass M passes through a wormhole mouth, "the entrance mouth has its mass increased by M, and the exit mouth has its mass reduced by an amount -M", and that this can eventually cause one of the mouths to have a net negative mass. Presumably light passing through a wormhole could have the same effect, based on mass-energy equivalence in general relativity. Pervect notes that in this context the "mass" being discussed is the ADM mass, and links to this post discussing the technical details of calculating the ADM mass of wormhole mouths, as well as some issues relating to quantum uncertainty in mass--I don't know enough general relativity to follow the technical details here, but the author seems to say that the mass of a mouth cannot actually become negative, perhaps for reasons relating to the "quantum inequalities" postulated to restrict negative energy that are discussed in this article. (maybe when John Cramer talked about the mass becoming negative he was giving the answer in "pure" general relativity without considering quantum physics?) Hopefully someone else who understands these topics better will weigh in, but I thought these links would be useful as pointers to research that would likely be relevant to answering your question.