I believe that there is a typo. Go back and count the zeros in the second capture you provided.
To answer your question, vfoA and vfoB are memory locations where the program is storing the values that the operator has selected. Depending on which one is active, that value is used to adjust the internal oscillator (Clock2 for main frequency) which is APPROXIMATELY 45MHz + the vfoA or vfoB depending on which is selected.
So if the vfoA = 7.100200 MHz (7100.200kHz)
Then the Clock2 VFO of the Si5351a will be set to approximately 7.1002 + 45 MHz = 52.1002 MHz
If the vfoB is 14.19735 MHz (14197.35 kHz) and selected then Clock2 of the Si5351a is approximately 59.19735 MHz.
Hope this helps.