Count resources in each category I have the following python script like:

query_parser.py

#!/usr/bin/env python


def count_resources_per_category(resource_names):
    APPS_MAP = {
        'billing': ["paypal", "checkout", "cart", "order-service"],
        'browse': ["content", "localization", "pricing", "search",],
        'some_category': ["random-1", "random-2"],
    }
    results_map = {}
    for key, _ in APPS_MAP.items():
        results_map[key] = 0
    results_map['other'] = 0
    # now we have:
    #   results_map = {
    #     'checkout': 0,
    #     'browse': 0,
    #     'MCM': 0,
    #     'other': 0
    #   }
    for resource_name in resource_names:
        FOUND_IT = False
        for key, app_names in APPS_MAP.items():
            if FOUND_IT:
                break
            for app_name in app_names:
                if app_name in resource_name:
                    results_map[key] += 1
                    FOUND_IT = True
                    break
        if not FOUND_IT:
            results_map['other'] += 1

    return results_map


def determine_if_resource_count_issue(results_map):
    RESULT_FOUND = False
    for key, count in results_map.items():
        if count:
            if RESULT_FOUND:
                print("The script isn't gonna work as planned")
                return True
            RESULT_FOUND = True
    print("All good :)")
    return False


#--------------------------------------------------------------------

""" TESTS """

# test_count_resources_per_category():
test_input = [
    'some-prefix-pricing', # browse
    'some-prefix2-cart', # billing
    'content', # browse
    'some-random-thing-not-found' # other
]
expectation = {
    'browse': 2,
    'billing': 1,
    'some_category': 0,
    'other': 1
}
result = count_resources_per_category(test_input)
assert result == expectation

#--------------------------------------------------------------------

# test_determine_resource_count_issue_passes():
test_input = {
    'browse': 234234234,
    'billing': 0,
    'some_category': 0,
    'other': 0
}
assert False == determine_if_resource_count_issue(test_input)

#--------------------------------------------------------------------

# test_determine_resource_count_issue_fails():
test_input = {
    'browse': 2,
    'billing': 1,
    'some_category': 0,
    'other': 1
}
assert True == determine_if_resource_count_issue(test_input)

#--------------------------------------------------------------------


It has some tests at the bottom explaining how it works, but the goal is to ensure each result from running a query is only for 1 category at a time.

I was wondering what the clean, pythonish way to handle these nested loops would be without the FOUND_IT and RESULT_FOUND flags. Any improvements welcome.
You don't need to initialize the keys of results_map to 0. You can do it on the fly. Here is a generic example:

counts = {}
for value in values:
    if value in counts:
        counts[value] += 1
    else:
        counts[value] = 1

Even better, use collections.Counter instead. Generic example:

from collections import Counter

counts = Counter()
for value in values:
    counts[value] += 1

Refactor the method to check if app_name is in resource_name. Put APPS_MAP outside the functions so it can be accessed as a global variable.

def get_category(resource_name):
    for key, app_names in APPS_MAP.items()
        for app_name in app_names:
            if app_name in resource_name:
                return key
    return 'other'

Now you can rewrite the loop in count_resources_per_category() like this:

from collections import Counter

results_map = Counter()
for resource_name in resource_names:
    results_map[get_category(resource_name)] += 1
return results_map

It seems like determine_if_resource_count_issue() is checking to make sure only one field is non-zero. Here's a quick one-liner replacement. Put the print() statements outside this function.

def determine_if_resource_count_issue(results_map):
    return [v != 0 for v in results_map.values()].count(True) == 1

get_category() as implemented above is computationally expensive. Depending on your use case, you may want to consider using regex.
Don't pad your question with lorem ipsum lol.