undefinsplit column by delimitered

I adapted the svcript of an older similar subject

let
    Source = Table.FromRecords(Json.Document(Binary.Decompress(Binary.FromText("i65Wcs7PKc3NM1SyUnIsKMjJTE4syczPs1IIS8wpTTXUUQhKLSxNLS6BChjpKLikFiQWleSm5sHEjHUUAhJLkjMy89Ldi/JLC6DCJkq1OsimY+gDGo5poxG6jcZKtbEA",BinaryEncoding.Base64),Compression.Deflate))),

    ac = Table.AddColumn(Source, "Custom", 
            each Record.FromList(
                 List.Transform(List.Alternate(Text.SplitAny(_[Column1],":,"),1,1,0),Text.TrimStart),
                
                List.Transform(List.Alternate(Text.SplitAny(_[Column1],":,"),1,1,1),Text.TrimStart)
                
                )
    ),

    fields=List.Union(List.Transform(ac[Column1], each  List.Transform(List.Alternate(Text.SplitAny(_,":,"),1,1,1),Text.TrimStart)))

in
   Table.FromRecords(ac[Custom],fields,MissingField.UseNull)

written in a perhaps more orderly manner:

let
    Source = Table.FromRecords(Json.Document(Binary.Decompress(Binary.FromText("i65Wcs7PKc3NM1SyUnIsKMjJTE4syczPs1IIS8wpTTXUUQhKLSxNLS6BChjpKLikFiQWleSm5sHEjHUUAhJLkjMy89Ldi/JLC6DCJkq1OsimY+gDGo5poxG6jcZKtbEA",BinaryEncoding.Base64),Compression.Deflate))),

  selFieldOrValues = (string, fOv) => List.Transform(
      List.Alternate(Text.SplitAny(string, ":,"), 1, 1, fOv), 
      Text.TrimStart
    ),
  ac = Table.AddColumn(
      Source, 
      "Custom", 
      each Record.FromList(selFieldOrValues([Column1], 0), selFieldOrValues([Column1], 1))
    ),
  fields = List.Union(List.Transform(ac[Column1], each selFieldOrValues(_, 1)))
in
  Table.FromRecords(ac[Custom], fields, MissingField.UseNull)