Bash does allow aliases to contain aliases but it has built-in protections against infinite loops. In your case, when you type lsc, bash first expands the alias to:
ls -Flatr --color=always
Since ls is also an alias, bash expands it to:
lsc -Flatr --color=always
lsc is an alias but, quite sensibly, bash refuses to expand it a second time. If there was a program named lsc, bash would run it. But, there is not and that is why you get command not found.
Addendum
It is different when lscR runs. lscR expands to:
ls -FlatrR --color=always
Since ls is an alias, this expands to:
lsc -FlatrR --color=always
Since lsc is an alias, this expands to:
ls -Flatr --color=always -FlatrR --color=always
Since ls has already been expanded once, bash refuses to expand it a second time. Since a real command called ls exists, it is run.
History
As noted by Schily in the comments, bash borrowed the concept of not expanding an alias a second time is borrowed from ksh.
Aside
Aliases are useful but not very powerful. If you are tempted to do something complex with an alias, such as argument substitution, don't; use a shell function instead.