Meaning of the data input pins of a DAC
I don't think page 10 is the best description of the operation of these bits.  You would be better served to look at the bottom left of page 11:


  DAC TRANSFER FUNCTION
  
  Each I and Q DAC provides complementary current
  output pins: IOUT(A/B) and QOUT(A/B), respectively. Note that QOUTA
  and QOUTB operate identically to IOUTA and IOUTB. IOUTA will provide a
  near full-scale current output, IOUTFS, when all bits are high (i.e.,
  DAC CODE = 1023), while IOUTB, the complementary output, provides no
  current. The current outputs of IOUTA and IOUTB are a function of both
  the input code and IOUTFS and can be expressed as 
  $$
I_{OUTA} = (DAC\;CODE/1024)\times I_{OUTFS}
  $$
  $$
I_{OUTB} = (1023 - DAC\;CODE)/1024\times I_{OUTFS}
  $$
  where: DAC CODE = 0 to 1023 (i.e., decimal representation)


Note that 0 to 1023 is 10 bits -- presumably set by the data input pins.  The paragraph on page 10 is describing the internal implementation (which may still be useful information for some applications).