The simple answer is that it isn't resolving in both orthogonal directions equally well. The horizontal dimension is the binocular dimension, and from looking at your animation, it looks to have about ~3 times the resolution. The horizontal banding, I'm pretty sure is not ringing, and is in fact, representative of additional information.
This article does a nice job of explaining what is being seen, and it has a picture, both of a simulated depiction of Io from one of the two telescope mirrors, and the combined image.
The thing you should notice from the above image is that the 8.4 m telescope picture is at the same resolution as the vertical axis of your animation.
What is going on with the interferometer image?
By increasing the separation between telescopes, we increase the angular resolution. Just about everyone with a casual interest in astronomy will have learned that fact in this past week.
But the other thing you do, is introduce interference patterns. The two telescopes essentially act like a double slit experiment.
In the vertical (non binocular) dimension, the diffraction can be treated like a single-slit experiment. The angular intensity formula for a single slit is as follows:
$$ I(\theta) = I_0 sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $b$ is the diameter of each mirror, and $\lambda$ is the wavelength of the light.
The horizontal (binocular) dimension acts as a double slit. The double slit formula is as follows:
$$ I(\theta) = I_0 cos^2(\frac{\pi d sin(\theta)}{\lambda})sinc^2(\frac{\pi b sin(\theta)}{\lambda})$$
where $d$ is the distance between the centres of the two telescopes.
When you combine these two together, using the telescopes dimensions that you quotes (b=8.4, d=14.4), you come up with a pattern remarkably close to what you actually see from the telescope.
On the left, screencapture from above animation, on the right, the predicted double slit intensity.
Fringe removal
It seems that the animation you saw is based off unprocessed images from the LBT. Obviously, they have methods for removing the fringe bands. As to how, I have no idea. I found a paper that discusses interferometry in depth and they mention:
What we therefore have is a series of fringes, whose amplitude is given by the Fourier transform of the source intensity distribution.
In practice, steps are usually taken to get rid of the fringes using a
phase rotation whose rate is known (as both B and s are known). This
is done in optical interferometers by use of accurate delay lines to
compensate for the path difference, and in radio interferometers by
the insertion of electronic delays.
But I have no idea what that means. Perhaps some boffin from Physics.SE would be nice enough to answer.
