Assigning a command to variable

The line you wrote defines a variable whose value is the string ls -f | grep -v /. When you use it unquoted, it expands to a list of words (the string is split at whitespace): ls, -f, |, grep, -v, /. The | character isn't special since the shell is splitting the string, not parsing it, so it becomes the second argument to the ls command.

You can't stuff a command into a string like this. To define a short name for a command, use an alias (if the command is complete) or a function (if the command has parameters and is anything more than passing some default arguments to a single command). In your case, an alias will do.

alias my_File='ls -f | grep -v /'
my_File

$ my_File='ls -f | grep -v '\/''
$ $my_File 
ls: cannot access |: No such file or directory
ls: cannot access grep: No such file or directory
[snip]

When interpreting $my_File, bash treats the characters in it as just characters. Thus, for one, the command line has a literal | character in it, not a pipe.

If you are trying to execute $my_File on the command line and have the pipes work, you need eval $my_File.