Error Handling is weird

[PythonAndArduino]

So, I have a try/except sequence to make sure my program doesn't get bogged by a user entering a non-number in response to a prompt asking for a number. Oddly enough, when it gets a non-number, it prints my message I have set up as the action for the "except" twice rather than once. Why is it repeating the try twice?

import random

def game (theNumber, newNumber):
    try:
        int(newNumber)
    except:
        print("That's not going to work. I need a number from 0 to 10.")
        return False, False, theNumber

    myNumber = int(newNumber)
    if myNumber >= 11 or myNumber < 0:
        print("I said 0 to 10. Let's try again:")
        return 
    elif theNumber == myNumber:
        return True, True, theNumber
    else:
        return True, False, theNumber

while True:
    
    doYou = input("Would you like to play? ")
    if doYou == "Yes":
        theNumber = random.randrange(0,10)
        newNumber = input("Guess a number from 0 to 10. ")
        game(theNumber, newNumber)
    else:
        print("Consent is vital, I need a clear answer: Yes")

    myResult = game(theNumber, newNumber)

    #print(myResult[0])
    #print(myResult[1])

    if myResult[0] == False:
        continue

    if myResult[1] == False:
        print("Nope, the answer was", theNumber)
    elif myResult[1] == True:
        print("Congratulations, you guessed it:", theNumber)
    else:
        continue

[Mekire]

You are calling game inside your if block and outside so all that code runs twice on yes and once on no:

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    doYou = input("Would you like to play? ")
    if doYou == "Yes":
        theNumber = random.randrange(0,10)
        newNumber = input("Guess a number from 0 to 10. ")
        game(theNumber, newNumber)
    else:
        print("Consent is vital, I need a clear answer: Yes")
 
    myResult = game(theNumber, newNumber)

It should instead be only called once, but let's look a little more at your input validation.  Firstly never catch a naked except.  In this case the exception you want is ValueError. 
Here is a look at a more elegant way to do this:

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def get_number():
    while True:
        try:
            user_input = int(input("Guess a number from 0 to 10: "))
            if not 0 <= user_input <= 10:
                raise ValueError
            break
        except ValueError:
            print("I said 0 to 10. Let's try again.")
    return user_input

This function will loop every time it gets an invalid input, instead of just looping the whole program.

This is an organized rewrite of what you have so far.  Obviously try to learn from it what you can rather than simply copy it:  

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import random


def get_number():
    while True:
        try:
            user_input = int(input("Guess a number from 0 to 10: "))
            if not 0 <= user_input <= 10:
                raise ValueError
            break
        except ValueError:
            print("I said 0 to 10. Let's try again.")
    return user_input
    
 
def game():
    the_number = random.randint(0,10)
    guess = get_number()
    win = guess == the_number
    if win:
        print("Congratulations, you guessed it: {}".format(the_number))
    else:
        print("Nope, the answer was {}".format(the_number))
    return win
        

def main():
    play = True 
    while play:
        do_you = input("Would you like to play (yes/no)? ").lower()
        if do_you == "yes":
            game()
        elif do_you == "no":
            print("Thanks for playing!")
            play = False
        else:
            print("Consent is vital, I need a clear answer (yes/no).")


if __name__ == "__main__":
    main()