capture groups
Your first example was close, but you're telling sed that you want to take 2 characters at a time with the \(.\)\(.\) bit, to do it using a capture group you could adjust it like so:
$ echo ahebhaaa | sed 's/\(.\)/\1\n/g' a h e b h a a a
using &
As to why the sed example with the & works. The & in a search and replace is what ever characters the regular expression matched. So everything that was not a \n.
Examples
$ echo "123 abc" | sed 's/[0-9]*/&&/' 123123 abc
The first & prints the characters that were matched by the pattern ([0-9]*), 123. The second & prints them a second time.
$ echo "123 abc" | sed 's/ab/&&&/' 123 abababc
The pattern we're looking for is ab. As we walk the string 123 abc sed is printing the non-matched characters, 123. Then the string ab is encountered which at that point matches what we're searching and replacing on. So sed then replaces it with 3 copies of what matched (abbab). Finally sed prints the c.
The notation [^\n] creates a set of not a end of line character. So think of sed as it's walking along your string of text, ahebhaaa, it's testing each of these characters and saying "Is this not a \n"? If it's not a end of line character, then sed does a search and replace on this character and the & prints what was matched, i.e. the character, along with a new line character. It then repeats this as it walks along the ahebhaaa string.
References
