Equivalence from multi-tape to single-tape implies limited write space?

Question

Suppose I have the following subroutine, to a more complex program, that uses spaces to the right of the tape:

$A$: "adds a $ at the beginning of the tape."

So we have: $$ \begin{array}{lc} \text{Before }: & 0101\sqcup\sqcup\sqcup\ldots\\ \text{After }A: & \$0101\sqcup\sqcup\ldots \end{array} $$

And it is running on a multi-tape turing machine. The equivalence theorem from Sipser book proposes we can describe any multi-tape TM with a single-tape applying the following "algorithm", append a $\#$ at every tape and then concat every tape in a single tape, and then put a dot to simulate the header of the previous machine, etc, etc.

With $a$ and $b$ being the content of other tapes, we have: $$ a\#\dot{0}101\#b $$ If we want to apply $A$ or another subroutine that uses more space, the "algorithm" described in Sipser is not enough, I can intuitively shift $\#b$ to the right, but how to describe it more formally for any other subroutine that uses more space in the tape? I can't figure out a more general "algorithm" to apply in this cases.

Answer 1

Sipser's book handles this case:

If at any point $S$ moves one of the virtual heads to the right onto a $\#$, this action signifies that $M$ has moved the corresponding head onto the previously unread blank portion of that tape. So $S$ writes a blank symbol on this tape cell and shifts the tape contents, from this cell until the rightmost $\#$, one unit to the right. Then it continues the simulation as before.