Note: this answer is for the first version of the question.
First, let's define
i with the tab:
$ i=$'ls -l\tcd ~'
Now, let's try your commands without and then with double-quotes:
$ echo $i | cut -f2
ls -l cd ~
$ echo "$i" | cut -f2
If you want
cut to work as expected, you need to put
$i in double-quotes. Without the double-quotes, the shell performs, among other things, word splitting which results in the tab being replaced by a blank. This prevents the
cut command from working as you expect.
Doing the assignment to Command does not change this:
$ Command=$(echo $i | cut -f2); declare -p Command
declare -- Command="ls -l cd ~"
$ Command=$(echo "$i" | cut -f2); declare -p Command
declare -- Command="cd ~"
You haven't provided the larger context here but, in general, it is a bad idea to try to put commands in variables. See "I'm trying to put a command in a variable, but the complex cases always fail!".
Separately, regarding the variable
COMMAND, it is better practice to use lower or mixed case for your shell variables. The system uses upper case for its variables and you don't want to accidentally overwrite one of them.