Time offset in a shift register
This is very simple if you think intuitively, A and B are the outputs of shift register and input x shifts bits at each rising edge of 2fs and frequency of x is half of 2f. In this question time offset means difference in time when first response reaches to A and B. For A it would reach at first rising edge and it would reach B in another two rising edge of 2fs so ultimately it is time period of x for 2fs frequency and time period of x (T offset)=1/fs. If you draw timming diagram correctly then you would get to this conclusion.
If still you have confusion then consider the time offset between geographical time zone sun rises first in newzealand and then in India and  the difference between them is called time offset the same analogy is there for shift register. At first rising edge of 2fs one bit of x would shift which is  at second edge it would reach to middle flip flop and at the 3rd edge it would reach B flop flop and between this interval one x completes it's one complete cycle hence time which x takes to complete cycle is 1/fs.
Please draw timing diagram as explained.