How many 12-2 cables can I put in a short run of conduit?
Liquidtight
If by &quot;LT&quot; you mean Liquidtight flexible conduit (metallic or nonmetallic), and you're looking for protection from physical damage. That's not an option, since it's not permitted to be used where subject to physical damage.
PVC
You'll have to use schedule 80 PVC, to provide protection from physical damage.
Calculate Fill
To calculate conduit fill for cables, you treat each cable as a single conductor and calculate the fill based on the major diameter of the cable (looked up from manufacturer documentation). For Southwire® Romex® SIMpull® 12/2 with ground cable, the major diameter is listed at 410 mils.
A = πr²
A = π * ((410/2)/1000)²
A = π * (0.205)²
A = π * 0.042025
A = 0.13202543126711106084639258818232 in.²
Since you're using the conduit as a sleeve, and not a complete conduit system, you should be able to fill the conduit to 100%. However, filling to 100% will make it impossible to pull the cables, so you'll want to stay well below that.
3/4&quot; Schedule 80 PVC has a total area of 0.409 in.², of which two 12/2 with ground NM-B cables will fill 65% of.  Depending on how long the sleeve is, this might make it quite difficult to pull the cables through.
Fill = Conductor Area / Total Area
Fill = (0.132025 * 2) / 0.409
Fill = 0.26405 / 0.409
Fill = 0.6456 = 65%
To put this in perspective. If this was two wires in a complete conduit system, you'd only be able to fill to 31% of the total area of the conduit (0.127 in.²).
Wet and Damp Locations
If this area can be considered damp or wet, you'll have to use cable rated for wet/damp locations.  In this case, you'll want to use UF-B cable (463 mil major diameter). Two of these cables will fill 82% of the conduit, making it nearly impossible to pull by hand (depending on the length).
Fill = (((π * ((463 / 1000) / 2)²) * 2) / 0.409) * 100
Fill = 82.33%