How does a CAN transceiver work?
Did I miss something or is my calculation valid (just for basic understanding :) )?  
  


First of all, the schematic shows BJT's. They have no drain source, but collector emitter.  

Next, the idea of calculation is in the right direction, however.
The datasheet does not directly give voltage drops across its internal components, but still defines values of \$V_{O(CANH)}\$ and \$V_{O(CANL)}\$ as well as the maximum differential voltage in the dominant state \$V_{DIFF(d)(o)}\$ (or: UT (Rtermination voltage) as you call it).



You're \$V_{O(CANH)}\$ and \$V_{O(CANL)}\$ might be valid, but the datasheet claims a UT (Rtermination voltage) = 3.8V will not occur as 3.0V is the maximum value.


  
  Can anyone explain how the recessive voltage generator in picture 2 works in detail? How big are the resistors to CANH and CANL?
  


The voltage source generates a voltage of (about) half the supply voltage (check parameter number D7 in the picture above). This voltage is fed to a buffer, which on its turn is enabled/disabled by the Power-On Reset block.
The values of those resistors are not given in the datasheet. The datasheet defines the resistance seen from the CAN nodes to ground in Figure 2-1.  



Parameter numbers D20 and D22 show the resistance is in the order of kWatts, oops, k&Omega; of course!


  
  Will Itermination double if there is a second termination resistor?  
  


By crude approximation: yes