Part of the problem

In your regex

oRE.Pattern = chr(34)+[0-9],[0-9]+chr(34)
                     ^

The + symbol after the chr(34) instructs the Regex engine to require one or more quotes

I think the + symbol should be moved to after the character set.

oRE.Pattern = chr(34)[0-9]+,[0-9]+chr(34)
                          ^

Proposed solution

You should capture the open/close quote internal digits into capture groups which are then used in the replacement.

Regex ("[0-9]+),([0-9]+")

Replace with: $1.$2

See this Live Demo

Sample input text:

20170927,Ford,,"6025,00",,"0,00",1,T
20170928,"Fiat, Opel",,"13587,17","13587,17",2,N

Sample after replacement

20170927,Ford,,"6025.00",,"0.00",1,T
20170928,"Fiat, Opel",,"13587.17","13587.17",2,N

Sample VB code

Note that the internal double quotes are doubled so that the string is constructed correctly.

VB Script

strTest = "20170927,Ford,,""6025.00"",,""0.00"",1,T" & vbCrLf & "20170928,""Fiat, Opel"",,""13587,17"",""13587,17"",2,N"    
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.Pattern = "(""\d*),(\d+"")"

strTest = re.Replace(strTest, "$1.$2")

VB.net

Dim sourcestring as String = "replace with your source string"
Dim replacementstring as String = "$1.$2"
Dim matchpattern as String = "(""[0-9]+),([0-9]+"")"
Console.Writeline(regex.Replace(sourcestring,matchpattern,replacementstring))

Regex Explanation

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    "                        '"'
--------------------------------------------------------------------------------
    [0-9]+                   any character of: '0' to '9' (1 or more
                             times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  ,                        ','
--------------------------------------------------------------------------------
  (                        group and capture to \2:
--------------------------------------------------------------------------------
    [0-9]+                   any character of: '0' to '9' (1 or more
                             times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    "                        '"'
--------------------------------------------------------------------------------
  )                        end of \2

Try this Code below:

Code:

strTest = "20170928,""Fiat, Opel"",,""13587,17"",""13587,17"",2,N"    'Store the text to be tested in this variable
Set re = New RegExp
re.Global=True
re.Pattern = "(""\d*),(\d+"")"
Set objMatches = re.Execute(strTest)

For Each match In objMatches
    strTest = Replace(strTest,match.Value,match.Submatches.Item(0)&"."&match.Submatches.Item(1))
Next

MsgBox strTest

Output:

Regex used:

("\d*),(\d+")

Click here for Regex Demo

Regex Explanation:

• ("\d*) - matches a " followed by 0+ occurrences of a digit. Parenthesis captures this whole part in group 1.
• , - matches , literally
• (\d+") - matches 1+ occurrences of a digit followed by ". Parenthesis captures this whole part in group 2.