Because you've simplified it to the point of being incorrect. First, you've confused big-O, which is an upper bound, with big-theta, which is a strict bound. But leaving that aside, big-O also drops constant factors, and your explanation doesn't. Also, it ignores finitely many 'irregularities', if I have a loop that normally runs n^2 times, except when n=17, 30, or 483, in which case it runs 10 billion times, that's still O(n^2). [That's actually implied by it ignoring constant factors, since you could just crank up the constant to 10 billion/17^2].

Finally, big-O notation describes arbitrary function growth, not just the runtime of some operation when applied to a list. It's used a lot in numerical analysis where you can show that some function can be approximated by, say, x - x^3/6 + x^5/120 + O(x^7).