Try matching:

((?:bPROB )?TEMPOb)

and replacing:

n1

See: regex101

Python code see online python:

# import the re module
import re

s="PROB TEMPO BKN001 TEMPO BKN009 PROB TEMPO BKN008 TEMPO BKN012"

# match and substitution pattern
mPattern=r"((?:bPROB )?TEMPOb)"
sPattern=r"n1"

# substitution; notice thta `.lstrip()` removes the leading `n`.
# dont know if thats desired. else just delete
res=re.sub(mPattern,sPattern,s).lstrip()

# result
print(res)

Outputs:

PROB TEMPO BKN001 
TEMPO BKN009 
PROB TEMPO BKN008 
TEMPO BKN012

Explanation

match:

• (): capture to group 1
• (?:bPROB )?: maybe preceeded with word PROB
• TEMPOb: the word TEMPO

replace:

• n1: linebreak and matched

You can use

import re

text = "PROB TEMPO BKN001 TEMPO BKN009 PROB TEMPO BKN008 TEMPO BKN012..."
print( re.sub(r"b(?!^)(PROBs+TEMPO|(?<!PROBs)TEMPO)bs*", "n\g<0>", text) )

See the Python demo and the regex demo. Output:

PROB TEMPO BKN001 
TEMPO BKN009 
PROB TEMPO BKN008 
TEMPO BKN012...

The b(?!^)(PROBs+TEMPO|(?<!PROBs)TEMPO)bs* regex matches any PROB TEMPO or TEMPO (not immediately preceded with PROB and a whitespace) a whole words.

If you need to support more than one whitespace before TEMPO, you’d need to intall PyPi regex module and then import regex as re and use b(?!^)(PROBs+TEMPO|(?<!PROBs+)TEMPO)bs*.

Although OP is asking about how to use RE for this it’s not clear if RE is actually a requirement. Why not use a simple loop?

text = "PROB TEMPO BKN001 TEMPO BKN009 PROB TEMPO BKN008 TEMPO BKN012"

p, *tokens = text.split()

result = p

for t in tokens:
    if t == "PROB" or (t == "TEMPO" and p != "PROB"):
        result += r"<br> "
    else:
        result += " "
    result += (p := t)

print(result)

Output:

PROB TEMPO BKN001<br> TEMPO BKN009<br> PROB TEMPO BKN008<br> TEMPO BKN012

Note:

Although not specifically stated in the OP, my guess is that br would not be required at the start of the output string.

This implementation for the relatively small input string is significantly faster than a RE solution

There are two ways to solve your problem.
One is to simply substitute both patterns “PROB” and “TEMPO” (not preceded by “PROB”) in turn:

import re

text = "PROB TEMPO BKN001 TEMPO BKN009 PROB TEMPO BKN008 TEMPO BKN012..."

new_text = re.sub(r"PROB", r"<br/> PROB", text)
new_text = re.sub(r"(?<!PROB) TEMPO", r"<br/> TEMPO", new_text)
print(new_text)

Will print:

<br/> PROB TEMPO BKN001<br/> TEMPO BKN009 <br/> PROB TEMPO BKN008<br/> TEMPO BKN012...

The (?<!PROB) part is a negative lookbehind as explained in the documentation and will match something not preceded by “PROB”.

The other solution uses two more regex tools: the or operator and backreferences and is a bit more elegant from the regex point of view.

import re

text = "PROB TEMPO BKN001 TEMPO BKN009 PROB TEMPO BKN008 TEMPO BKN012..."

new_text = re.sub(r"PROB|(?<!PROB) TEMPO", r"<br/> g<0>", text)
print(new_text)

Will print:

<br/> PROB TEMPO BKN001<br/>  TEMPO BKN009 <br/> PROB TEMPO BKN008<br/>  TEMPO BKN012...

The pattern to match is “PROB” OR “TEMPO” (not preceded by “PROB”).
The g<0> in the replacement string is a backreference to the entire match.

In your question, you attempted to use a backreference with started your replacement string with BKN0dd.
The intention is obvious but this will only resolve to literal backspace-d’s.
Backreferences tell the regex engine “that’s what a previous pattern matched”.

Also note that I used <br/> (not </br>) as this is the correct XHTML element for a line break.
This assumes your string is somehow used in an HTML document.
Wiktor’s answer demonstrates how to add actual line breaks.