How to find left-hand side of tape on a Turing Machine?
for above example before you start your tape is like this ($e$ means empty and right side of $&gt;$ is block of tape that you are reading it)

e&gt;001001e


as you said you should first mark $X$ on the first $1$ and it become like this

e00&gt;X001e


now you should clear two zeros because the number of zeros is twice of the number of ones. you can clear the first two zero of the tape. go to the beginning of the tape like this (when you read $e$ it means you are at the beginning of tape and go one step to the right so that marker be over the first symbol of tape)

e&gt;00X001e


now you can mark two zeros with $X$ or $Y$ it does't matter but for readability it's better to mark it with another symbol 

eY&gt;YX001e


now go back to the beginning of tape

e&gt;YYX001e


now it's like a for loop you should do this until there is no $1$ on the tape. when you got end of the tape while searching for $1$ it means that you ran out of ones. you should check two cases


is number of ones less than twice of zeros
is number of ones greater than twice of zeros


if the first case happens while you are searching for zeros you cant find two of them and you got to the end of tape like below

e&gt;00010111e


become

eYYYXYXY1&gt;e


so in this case you should reject the string.

and checking the second case is in the last step when you ran out of ones you should check that is there any zeros one the tape right now. if you found any zero you should reject.

remember never write empty over $0$ or $1$ because when you reach empty it means you are at the end or the beginning of the tape.

based on definition of Turing Machine you can run off the left hand side or not. there are Turing Machines that their tapes are infinite at the right side and finite at the left side so you can run off at the left side. but standard Turing Machine is infinite at both site so you can run off at the left hand side.