Describe a Turing machine that accepts a word iff there are twice as many 0's as 1's
Roughly speaking, the machine could act as follows:


Scan the input, looking for a 1 entry.
If no 1 entry is found, check that there are no 0 entries left, and halt at an accepting or rejecting state accordingly.
If a 1 entry is found, erase it.
Look for a 0 entry. if none is found, halt at a rejecting state. Otherwise, erase it.
Repeat step 4 one more time.
Go back to step 1.


When erasing an entry, you should change it to a symbol different from the blank symbol, so that you will be able in the future to scan through the entire breadth of the work.

If you have multiple tapes, you could try the following algorithm instead:


Scan the input. Whenever you see a 0, copy it to auxiliary tape 0. Whenever you see a 1, copy it to auxiliary tape 1.
As long as there are 1s in auxiliary tape 1: erase a 1, and erase two 0s from auxiliary tape 0; if there are not enough 0s, reject.
Accept iff there are no 0s in auxiliary tape 0.


This is a more efficient variant of the first algorithm.