SubString problems could be complicated, once one drills a bit in them. E.g., in the OP example, the substring KKYKDKSK also is a correct substring of KK*K, thus, it probably could be color coded as well.
In general, with some limitations the task, like searching for non-overlapping substrings and considering that the substring is present once per string, this is possible:

With some hardcoding of the variables and checking only for KK*K, this is how the main method looks like:
Option Explicit
Sub TestMe()
Dim myRange As Range: Set myRange = Worksheets(1).Range("A1:A2")
Dim myCell As Range
For Each myCell In myRange
myCell.Font.Bold = False
Dim subString As String
subString = findTheSubString(myCell.Value2, "KK*K")
Debug.Print myCell.text, subString
ChangeTheFont subString, myCell, vbBlue
Next myCell
End Sub
The function findTheSubString() takes the 2 strings and returns the substring, which is to be color-coded later:
Public Function findTheSubString(wholeString As String, subString As String) As String
Dim regEx As Object
Dim inputMatches As Object
Dim regExString As String
Set regEx = CreateObject("VBScript.RegExp")
With regEx
.Pattern = Split(subString, "*")(0) & "[\s\S]*" & Split(subString, "*")(1)
.IgnoreCase = True
.Global = True
Set inputMatches = .Execute(wholeString)
If regEx.test(wholeString) Then
findTheSubString = inputMatches(0)
Else
findTheSubString = "Not Found!"
End If
End With
End Function
The last part is to change the font of a specific substring in Excel range, thus the arguments are a string and a range:
Sub ChangeTheFont(lookFor As String, currentRange As Range, myColor As Long)
Dim startPosition As Long: startPosition = InStr(1, currentRange.Value2, lookFor)
Dim endPosition As Long: endPosition = startPosition + Len(currentRange.Value2)
With currentRange.Characters(startPosition, Len(lookFor)).Font
.Color = myColor
.Bold = True
End With
End Sub
.Executereferring to here? This half looks like a Regular Expression-type approach, without any RegExp object (which in any case would not be available on a Mac)trueif a match was found andfalseif not. It has nothing comparable toregexwhich is able getting the matching string parts too. So either you will usingregexor you needs finding your matches another way. InStr function would be a good start then.CreateObject("VBScript.RegExp")?Likeoperator in this question indicates a variable length pattern. Which is it?