Suppose root is your root node from the ElemenTree.
The namespace is read from the 'edgar:xbrlFiling' node's attribute 'xmlns:edgar':
xmlns:edgar="http://www.sec.gov/Archives/edgar"
ElemTree encodes edgar:any_tag as the python string:
ns + 'any_tag'
Where ns is the python string below:
ns = '{http://www.sec.gov/Archives/edgar}'
So to find all the xbrlFile nodes you can use the following XPath expression:
xbrlFiles = root.findall('.//'+ns+'xbrlFile')
To get the URL attribute you need to extract the ns+'url' attribute (in this case for the second file):
myurl = xbrlFiles[1].attrib[ns + 'url']