You haven't really asked a specific question but here are some comments that hopefully help you understand what's going on:
If you just enter
into your terminal the shell
expands the variable called
$PATH and then
attempts to execute its
contents which, obviously, doesn't make much sense with regards to
$PATH. Try running
and you will understand what happens. Note though that in most cases you do not want to store a command in a variable, you will want to read BashFAQ/050 aka I'm trying to put a command in a variable, but the complex cases always fail! to learn a little bit about the edge cases and why this is generally a bad idea unless you know exactly what you are doing.
echo $var you are printing the contents of the variable
$var to the screen. Note that
word splitting can occur and usually you will want to quote your variables to avoid this:
$ echo "$var"
This does not work because
cd is a command (OK, a shell builtin usually) and
.. is its parameter and the two need to be separated by whitespace so the shell's parser can pick these up as separate tokens:
$ cd ..
Note that you can set up an alias to make
$ alias cd..='cd ..'
but I would advise against that and learn how to properly use a Unix shell instead of making it mimic