Duck, duck, gone!

JavaScript (ES9), 227 bytes

This is similar to the Node version below but uses a formula based on parseInt() instead of Buffer() to identify the input verse.

This is ES2018 (aka ES9) because we're using a regular expression with the /s flag (dotAll).

s=>'Mother duck herself1and all23,,Three4two3,Five4four3,Four4three3,One01but none23,Two4one0'.split`,`[parseInt(s,30)&7].replace(/\d/g,n=>[x=' little duck',y=/ w.*\n/s.exec(s),' of the',x+='s',x+y+'but only '][n])+s.slice(-11)

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How?

In this version, we parse the entire input verse as base 30 (0 to t) and perform a bitwise AND with 7. The parsing stops on the first invalid character, leading to:

 verse | valid part | base 30 -> decimal | AND 7
-------+------------+--------------------+-------
   0   |  'fi'      |            468     |   4
   1   |  'fo'      |            474     |   2
   2   |  'three'   |       23973734     |   6
   3   |  't'       |             29     |   5
   4   |  'one'     |          22304     |   0
   5   |  'mother'  |      554838747     |   3

---

JavaScript (Node.js),  233 231  227 bytes

Saved 2 bytes thanks to @Shaggy

s=>'Three4two3,Four4three3,Mother duck herself1and all23,One01but none23,,,Two4one0,,Five4four3'.split`,`[Buffer(s)[2]%9].replace(/\d/g,n=>[x=' little duck',y=/ w.*\n/s.exec(s),' of the',x+='s',x+y+'but only '][n])+s.slice(-11)

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How?

The third character of each input verse can be used as a unique identifier. By taking its ASCII code modulo 9, we get:

 verse | 3rd char. | ASCII code | MOD 9
-------+-----------+------------+-------
   0   |    'v'    |     118    |   1
   1   |    'u'    |     117    |   0
   2   |    'r'    |     114    |   6
   3   |    'o'    |     111    |   3
   4   |    'e'    |     101    |   2
   5   |    't'    |     116    |   8

The output verses are encoded with the following templates:

 verse | template
-------+---------------------------------
   0   | 'Five4four3'
   1   | 'Four4three3'
   2   | 'Three4two3'
   3   | 'Two4one0'
   4   | 'One01but none23'
   5   | 'Mother duck herself1and all23'

Where each digit is replaced with a string according to the following table:

 digit | replaced with
-------+---------------------------------------------------
   0   | ' little duck'
   1   | / w.*\n/s.exec(s)
   2   | ' of the'
   3   | ' little ducks'
   4   | ' little ducks' + / w.*\n/s.exec(s) + 'but only '

Where the regular expression / w.*\n/s extracts this common part from the input:

  went out one day,[LF]
 over the hills and up away.[LF]
 Mother Duck said, "Quack Quack Quack Quack",[LF]

We finally add the last 11 characters of the input, which is " came back.".

Python 3, 267 263 254 bytes

4 bytes saved thanks to @ovs

def f(s):
 for a in zip(T[2:]+T,T):s=s.replace(*a)
 return s
T="8:9:and allHO1BnoneHT2No1T3Nt2F4Nt3FiveINf4MotherD herself"
for r in "H of theI,4ourI,3hreeI,2woI,1neL:,ILs:,L littleD,D duck,NBonly ,Bbut ".split(','):T=T.replace(r[0],r[1:])
T=T.split(':')

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Works by replacing the relevant parts by the respective parts of the next verse.

After the preinitialisation, T is ['8', '9', 'and all of the little ducks', 'One little duck', 'but none of the little ducks', 'Two little ducks', 'but only one little duck', 'Three little ducks', 'but only two little ducks', 'Four little ducks', 'but only three little ducks', 'Five little ducks', 'but only four little ducks', 'Mother duck herself'].

Alternative Python 2, 252 bytes

by @ovs

lambda s:reduce(lambda s,a:s.replace(*a),zip(T[2:]+T,T),s)
T="8:9:and allHO1BnoneHT2No1T3Nt2F4Nt3FiveINf4MotherD herself"
for r in "H of theI,4ourI,3hreeI,2woI,1neL:,ILs:,L littleD,D duck,NBonly ,Bbut ".split(','):T=T.replace(r[0],r[1:])
T=T.split(':')

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QuadR, 257 242 bytes

-14 thanks to Black Owl Kai, -1 thanks to Kevin Cruijssen

ive
Four
hree
Two( little duck)s
One little( duck)
Mother( duck) herself
four
two( little duck)s
only on(e little duck)
but none
and all of the
our
Three
wo
One\1
Mother\1 herself
Five little\1s
three
one\1
none of th\1s
and all
but only four

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