A one-liner. No reason to do this.

with open('x.py') as f: s = f.read()

read_file accepts a file-like object, read accepts a file name.

The main issue with read is that if the file doesn't exist, it fails silently, leaving the ConfigParser empty.

config_parser.read("config.ini")  # object is unchanged even if file doesn't exist

That cannot happen with read_file since you control the way the file is opened.

with open("config.ini") as f:
   config_parser = configparser.RawConfigParser()
   config_parser.read_file(f)

The other useful case is when you have a list of lines representing a .ini file contents.

read_file kind of "accepts" a string as input (since strings are iterable) but the result isn't useful:

>>> config_parser.read_file("config.ini")
Traceback (most recent call last):
  File "<string>", line 301, in runcode
  File "<interactive input>", line 1, in <module>
  File "L:\Python34\lib\configparser.py", line 690, in read_file
    self._read(f, source)
  File "L:\Python34\lib\configparser.py", line 1057, in _read
    raise MissingSectionHeaderError(fpname, lineno, line)
configparser.MissingSectionHeaderError: File contains no section headers.
file: '<???>', line: 1

This code will read the entire file into memory and remove all whitespace characters (newlines and spaces) from the end of each line:

with open(filename) as file:
    lines = [line.rstrip() for line in file]

If you're working with a large file, then you should instead read and process it line-by-line:

with open(filename) as file:
    for line in file:
        print(line.rstrip())

In Python 3.8 and up you can use a while loop with the walrus operator like so:

with open(filename) as file:
    while line := file.readline():
        print(line.rstrip())

Depending on what you plan to do with your file and how it was encoded, you may also want to manually set the access mode and character encoding:

with open(filename, 'r', encoding='UTF-8') as file:
    while line := file.readline():
        print(line.rstrip())