double quotes in zsh parameter expansion

Question

I have two test cases in zsh

1. A. without quotes

   ~$ y=(${(f)$(echo -e "a b\nc d")}); printf "<%s>\n" "${y[@]}"
   <a b c d>
   

2. B. with quotes

   ~$ y=(${(f)"$(echo -e "a b\nc d")"}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

However if I first assign the output of echo to a variable, the quotes do not have any effect:

1. C. without quotes

   ~$ x=$(echo -e "a b\nc d"); y=(${(f)${x}}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

2. D. with quotes

   ~$ x=$(echo -e "a b\nc d"); y=(${(f)"${x}"}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

Questions:

1. comparing A and B, what causes the differences?
2. comparing A and C, what causes the differences?

Answer 1

After some experiment, I feel the following rules may have been applied

Sec1. direct use

1. A. without quotes

   ~$ y=(${(f)$(echo -e "a b\nc d")}); printf "<%s>\n" "${y[@]}"
   <a b c d>
   

   unquoted $() produces words split by IFS: see http://zsh.sourceforge.net/Doc/Release/Expansion.html#Command-Substitution

2. B. with quotes

   ~$ y=(${(f)"$(echo -e "a b\nc d")"}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

   quoted $() produces a single string

Sec2. assignment to scalar

1. C. without quotes

   ~$ x=$(echo -e "a b\nc d"); y=(${(f)${x}}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

   when being assigned to a scalar, it works as automatically quoted

2. D. with quotes

   ~$ x=$(echo -e "a b\nc d"); y=(${(f)"${x}"}); printf "<%s>\n" "${y[@]}"
   <a b>
   <c d>
   

   when being assigned to a scalar, it works as automatically quoted

Sec3. assignment to array

y=($(echo -e "a b\nc d"))

Follows Sec1 http://zsh.sourceforge.net/Doc/Release/Expansion.html#Command-Substitution