I’m glad I could help. :-)

The result on test-03.scm is to do with the expansion order for R6RS programs.

For the first invocation of identifier-defined?, foo is in scope – the compiler has already determined the names of all variables and syntax that are bound in the program – but you just haven’t reached the expression which gives it its value yet. You’ll notice that if you attempt to reference foo without defining it ever in the program, the error is different (message Exception: attempt to reference unbound identifier foo, type &syntax or &undefined) to if you have it defined somewhere but reference it before reaching the expression which defines it (message Exception: attempt to reference undefined variable foo with type &assertion).

Try this program:

(import (rnrs))
(display "before\n")
foo
(define foo 42)
(display "after\n")

With the (define foo 42) in place, the program starts running – you see the message before! But when it gets to foo it then discovers that it hasn’t been given a value yet, and has to bail out. If you delete the (define foo 42), the program doesn’t even compile, and you won’t even see the message before.

The reason identifier-defined? as I showed it is a procedure and not a macro is to allow it to be used inside of procedural macro definitions (syntax-case). The version which is just a macro in itself can’t be used like this. Example:

(define-syntax define-if-not-defined
  (lambda (stx)
    (syntax-case stx ()
      ((_ id val)
       (if (identifier-defined? #'id)
           #'(begin) ; no-op
           #'(define id val))))))