Indentation-based Sort

Python 2, 117 bytes

lambda S:[s[-1]for s in sorted(reduce(lambda t,s:t+[[v for v in t[-1]if v.count(' ')<s.count(' ')]+[s]],S,[[]]))[1:]]

Try it online!

Takes as input a list of strings; and outputs a list of strings, sorted as required.

The idea is to turn each element into a list which contains the "absolute path" as a list; and then let Python handle the sorting. E.g. if the input is:

[
 'a',
 ' c',
 '  d',
 ' b'
]

Then via the reduce(), we convert to a list of lists:

[
 ['a'],
 ['a',' c'],
 ['a',' c', '  d'],
 ['a',' b']
]

which gets sorted as:

[
 ['a'],
 ['a',' b']
 ['a',' c'],
 ['a',' c', '  d'],
]

and then output the last element of each list in the list-of-lists to get:

[
 'a',
 ' b'
 ' c',
 '  d',
]

APL (Dyalog Unicode), 31 bytes^SBCS

Anonymous prefix lambda, takes and returns list of strings.

{⍵[⍋{1=≢⍵:⍺⋄⍵}⍀↑' '(,⊂⍨⊣=,)¨⍵]}

Try it online!

{…} "dfn"; ⍵ is argument

 ⍵[…] index the argument with the following indices:

  ' '(…)¨⍵ apply the following tacit function to each string with space as left argument:

   , concatenate the space to the string

   ⊣= Boolean list indicating where the space is equal to each character that

   ,⊂⍨ use that to partition (begin part where true) the concatenation of space and string

  ↑ mix list of lists of strings into matrix of strings

  {…}⍀ vertical cumulative reduction by this "dfn"; ⍺ and ⍵ are upper and lower args:

   ≢⍵ the length of the lower string

   1= is that equal to 1? (i.e. is there nothing but the single space there?)

   :⍺ if so, return the upper argument

   ⋄⍵ else, return the lower argument

  ⍋ grade up (find indices which will sort that)

Retina, 47 bytes

+-1m`(?<=^\2(\S+).*?¶( *)) 
$1 
O$`
$L$&
\S+ 
 

Try it online! Note: Several lines have trailing spaces. Explanation:

+-1m`(?<=^\2(\S+).*?¶( *)) 
$1 

The first step is to insert each word into following lines at the same indentation. For example, with the lines aisle, wasabi and elf the resulting lines are aisle, aisle wasabi and aisle wasabi elf. I discovered this regex by trial and error so there may be edge cases with it.

O$`
$L$&

We can now sort the lines case-insensitively.

\S+ 
 

Delete all of the inserted words.