It looks good. A little background...
The 6 main orbital elements describe the size, shape, and orientation of the orbit, as well as the object's position on that orbit at a particular time.
Sometimes mean anomaly is used for position. Sometimes time since (or to) periapsis passage is used.
So for example, imagine an orbit with a period of 60 minutes. If t0 is 15 minutes, that means that 1/4 of an orbit ago it was at periapsis.
So 15/60 or 1/4 of an orbit, which is 360 degrees is 90 degrees.
360-90 = 270. So the M is 270 degrees.
Notice I considered D to be 0 for simplicity. If D is after t0, you get a negative number, which is fine. You'll get an angle greater than 360, so you have to subtract off 360 to get back in the 0-360 range.
Note that they give semi-major axis in milli arcseconds. This needs to be converted to a distance in order for the conversions from orbital elements to cartesian coordinates to make sense. We need to know the distance from the Earth to the black hole to make the conversion.
from the paper
...If not specified otherwise (§ 3.3), we adopt throughout this paper R0 = 8 kpc for the Galactic center distance...
...The updated estimate of distance to the Galactic center from the S2 orbit fit is R0 = 7.62 ± 0.32 kpc...
So therefore, looking at the semi-major axis for S1 = 0.412 arcsec and converting to meters gives:
8000 parsecs x 3.08568025E+16 meters / pc * sin(0.412 / 3600) = 493075008240113 meters
As a double-check, we can plug into the period formula (P =2pi sqrt(a^3/(GM)) and convert to years.
To do this, we need to know the mass of the black hole.
...The best-fit central mass
9 for an assumed distance of 8 kpc is (4.06 ± 0.38) × 10^6 (solar masses)...
...The updated estimate of distance to the Galactic center from the S2 orbit fit is
R0 = 7.62 ± 0.32 kpc, resulting in a central mass value of (3.61 ± 0.32) × 10^6 (solar masses)...
...Measurements of stellar velocities and (partial) orbits have established a compelling case that this dark mass concentration is a massive black hole of about (3[PLAIN]http://cdn.iopscience.com/icons/EJ4/AJ/ucp-icons/ndash.gif4) × 10^6 (solar masses)...
I think they made a typo in that middle sentence. I think they meant the distance from Earth, not S2. That would put S2 about as far away as Earth.
The period formula is P = 2pi * sqrt(a^3/(GM))
This gives seconds. Dividing by 3.155 x 10^7 gives years.
2*pi*sqrt((8000*3.08568025E+16*sin(.412/3600))^3 / (6.67e-11*3.61e6*1.99e30))/3.155e7
gives 99.6 years. Using 4 instead of 3.61 gives 94.6 years, which is in rough agreement with the period given for S1 in the data you quoted.
You can copy and paste the above formula into Google,and play with various numbers. It outputs years.
You've now got everything properly converted for use in the calculator:
http://orbitsimulator.com/formulas/OrbitalElements.html
I'm looking forward to seeing your animation! Let me know if you need help.