Why doesn’t 0 divided by 0 equal 0?

Well the great thing about math is we can just pretend for a second it does, and see what happens. If something goes catastrophically wrong, we know we made a bad assumption. So let’s see what happens.

Clearly, 1+0=1. So if 0/0=0, then 1+0/0=1. Now we can add fractions by finding a common denominator, so (0x1+0)/0=1. The left side is 0/0 (which is 0) but the right side is 1. So some assumption we made doesn’t work, and the only thing we really assumed was 0/0=0. So that can’t be true.
By this line of reasoning, isn’t 7/7 also zero?

If you’re using the word “add” to talk about division you’re thinking about it wrong. Division is the inverse of multiplication, it has nothing to do with addition.

Division, as the inverse of multiplication, can be thought of as a question. That question is “what do I need to multiply this by to get that?”. For example, when you say “8/2”, you are asking what number you need to multiply 2 by to get 8, the answer of course being 4.

With 0/0, you are asking the question “what number do I need to multiply 0 by to get 0”? 0 would work. But so would 1, 2, π, any other number, because (almost by definition) any number multiplied by 0 is 0. 0/0 is the only “division question” with multiple answers, all just as good as each other without further context.

And in contrast when you ask the question “1/0” (or 2/0, etc), there is no answer. No number yields 1 when multiplied by 0.
2/2=1

1/1=1

Wouldn't 0/0 = 1?

It goes in there once, with no remainder.

(This is facetious btw)
If you study calculus you will find that most of the interesting limits happen when they are at first evaluated as 0/0.  After factoring out the factor that makes it zero we can find an actual value that has geometric and physical meaning, showing that 0/0 is often obscuring some other value which hides behind the veil
Your logic is correct. You're implementing it wrong.

9/3 = 3 (3 added 3 times will give 3; 3 goes 3 times in 9)

12/6 = 2 (6 added 2 times will give 12; 6 goes 2 times in 12)

So 0/0 = ?

Lets go by the same logic (your logic). How many times zero must be added to get zero? Well,

zero added zero times = zero

Zero added 2 times = zero

Zero added infinite times = zero

So the answer seems to be anything. Its undefined.
a/b= c is equivalent to a= bc.

Yes, it is true that 0= 0(0) so shouldn't we say that 0/0= 0?

But it is also true that 0= 1(0) so shouldn't we say that 0/0= 1?

And it is true that 0= 2(0) so shouldn't we say that 0/0= 2?

In fact any number, a, satisfies 0= a(0) so shouldn't we way 0/0= any number?
the definition of a/b is a * b^(-1), where b^(-1) means the number such that when you multiply it by b, the result is 1. if b = 0, there is no such number, so 0^(-1), and hence also a/0, are undefined.
0 goes into 0 two million five hundred twenty-four thousand eight hundred eighty-six times.  I mean, 2524886·0 = 0, right?
As the inverse of multiplication, a division of a/b can be thought as "how many b would make a to zero if we subtract them". For example, if we have 6/2, it can be viewed as 6 - 2 - 2 - 2 = 0 or 6 - 2(3) = 0. By 0/0, you get this equation 0 - n0 = 0. So what is the value of n? 1? 4201? -29103? It all satisfies the equation, thus undefined.
So there are three cases of interest here:

1. 5/5 = 7/7 = -13/-13 = 1; "same thing/same thing" is 1,
2. 0/5 = 0/7 = 0/-13 = 0; 0 divided by something is 0, and
3. 5/0, 7/0, -13/0 are all undefined; something divided by 0 is undefined.

So where does 0/0 properly fit?  It appears to fit all three cases, so how would you choose in a way that wasn't arbitrary?  Which should be more important?

"Oh, but Professor shellexyz, it depends on whether the 0 in the numerator is fixed and you just change the denominator.  Or if the numerator and denominator are always the same.  You want it to be consistent."

Yes, consistency is important, and the fancy word for what you want is actually "continuity".  You've discovered the idea of limits.  Let's talk about that, at length, in gory detail, with many, many examples over the next two weeks.

Welcome to differential calculus.