In other words (I thought) /(b*)/ stops after the first failure, at the start of the string,

s/failure/success/

whereas adding the /g would tell the regex engine to keep on trying until it reaches the end of the string.

What /g does on a regex depends on context. In boolean scalar context it matches once, and stores the position in pos. If you execute it a second time, it starts off from where it left.

The background is that you can write

while (/(b*)/g) {
...
}
[download]

and get a new match for each iteration:

$perl -e '$_="abbbabc";
while (/(b*)/g) { print "($1)\n" }'
()
(bbb)
()
(b)
()
()
[download]

Update: Answer to the second question

That this code produces two replacements for the string of four 'b's remains a puzzle. Why does this appear (this may be my error) that the regex conflates two 'b's rather than all four?

A naiive substitution implementation would loop on s/b*/^/, because it would continue to replace the empty string with ^ forever on.

Perl is a bit more sophisticated: It detects a zero-width match, and before doing a second substition of a zero-width match at the same position it bumps along, and tries in the next position.

So applying s/b*/^/ on abba make these steps:

abba
| match zero b's before a
^abba
 | match zero b's again. Don't substitute here, bump along
^abba
  | match 'bb'
^a^a
   | match zero b's
^a^^a
    | match zero b's, don't substitute but bump along
^a^^a^
     | match zero b's, don't substitute but bump along
[download]

You can watch it work; I didn't find a way to get the modified string, but at least you can monitor the match positions:

$ perl -le '$_ = "abba"; s/b*/print pos; "^"/eg; print'
0
1
3
4
^a^^a^
[download]