In the book you mentioned above you simulate a infinitely long bidirectional TM increasing the number of symbols and by "“folding” it in an arbitrary location". Thus each cell in the unidirectional TM will represent 2 symbols or a pair. So this can be x/(symbol or blank), x/y or (symbol or blank)/x. Now T(n) is the worst case scenario. Let us assume that the current state of the TM is as follows
^ symbol for blank.
e/h|l/^|l/^|o/^|!/^|?/^ (this is equivalent to h|e|l|l|o|!|? in a bidirectional TM, basically bend on the first position itself). with head pointing to ?/^, now if I want to write to the position I have to write as a pair. But I cannot read the second symbol in the pair without iterating to the left most position and then to the right again. So from the head position ?/^ we have to iterate 6 times to the left and 5 times to the right to read "^". Once we read "^" we have to return, so iterate back so 6 times to the left and then 5 times to the right. Now that we know the second symbol in the pair, we can write the new symbol for ex . "u/^". In a more generalized sense to write something in a Tape in the above state we have iterate the length of the tape(n) 4 times. So it will take 4 times more time to halt on a regular TM in the worst case, so t(n) = 4t'(n).