Your example code is wrong. This works:

import datetime

datetime.datetime.strptime("21/12/2008", "%d/%m/%Y").strftime("%Y-%m-%d")

The call to strptime() parses the first argument according to the format specified in the second, so those two need to match. Then you can call strftime() to format the result into the desired final format.

If you split an empty value, it won't have anything at index 1. You can use an if statement to check this:

for row in updateCursor:

    theWrongDate = row.RPT_DTE
    theRightDateList = theWrongDate.split("/")

    if len(theRightDateList) == 3:
        year = '19' + str(theRightDateList[0])
        month = theRightDateList[1]
        day = theRightDateList[2]
        theNewDate = "{0}/{1}/{2}".format(month, day, year)

        # verify I have the next couple lines correct
        row.setValue(RPT_DTE, theNewDate)
        updateCursor.updateRow(row)

You can use the parse_dates and dayfirst arguments of pd.read_csv, see: the docs for read_csv()

df = pd.read_csv('myfile.csv', parse_dates=['Date'], dayfirst=True)

This will read the Date column as datetime values, correctly taking the first part of the date input as the day. Note that in general you will want your dates to be stored as datetime objects.

Then, if you need to output the dates as a string you can call dt.strftime():

df['Date'].dt.strftime('%d/%m/%Y')

Just put that into the str(...) function:

import datetime
my_date = datetime.date(2022, 6, 7)
print(str(my_date))  # prints 2022-06-07

Technically, you can just print it and not make it a string first. But putting it in str means that instead of printing it, you could save that string to a variable.

If you need more advanced formatting options, then you can do what @FObersteiner suggested. But the format you want happens to be the default, so this will do if you just want that one format