Calling the Value of a Function from within a VBA Sub - vba

Question

Asking users to identify the folder where they want certain files to be saved, then save new workbooks to that folder. Sub calls the Function that sets this folder path to JobFolder. When calling JobFolder folder path from sub, the sub runs the Function all over again. Just need the value of JobFolder. Thank you.
Sub ExportJobFiles()
'Ask user to set job folder
Dim JobFolder As String
JobFolder = GetFolder()
'Ask user to set suffix for file names
Dim FileNameSuffix As Variant
Dim Default As String
Default = Worksheets("Summary").Range("E2").Value
FileNameSuffix = InputBox("Input suffix for job files", , Default)
'Creates job files in job folder
Dim SummaryFileName As String
Dim AFileName As String
Dim BFileName As String
Dim CFileName As String
SummaryFileName = JobFolder & "/Summary_" & FileNameSuffix & ".xls"
AFileName = JobFolder & "/A_" & FileNameSuffix & ".xls"
BFileName = JobFolder & "/B_" & FileNameSuffix & ".xls"
CFileName = JobFolder & "/C_" & FileNameSuffix & ".xls"
Workbooks.Add.SaveAs FileName:=SummaryFileName
Workbooks.Add.SaveAs FileName:=AFileName
Workbooks.Add.SaveAs FileName:=BFileName
Workbooks.Add.SaveAs FileName:=CFileName
End Sub
Function GetFolder() As String
With Application.FileDialog(msoFileDialogFolderPicker)
.Title = "Select a Folder"
.AllowMultiSelect = False
.InitialFileName = ""
.Show
GetFolder = .SelectedItems(1)
End With
End Function

Answer 1

Call JobFolder
That's calling the function (with an obsolete explicit call syntax) and discarding its return value.
Basically, that line does nothing useful.
Instead, declare a local variable:
Dim folder As String
And assign it with the function's return value:
folder = JobFolder
As noted in comments, you should also avoid GoTo jumps in or out of With blocks.
If the With block was written like this:
With Application.FileDialog(msoFileDialogFolderPicker)
'...
End With
Then jumping out of it would leave the object reference held by the With block in limbo. The only reason that's not happening with your code is because you already have a reference to the With block variable outside the With block, and you're destroying it manually... which isn't necessary if you let the With block handle the reference.