No, that's only true at the limit as n goes towards infinity.

And even if you see it as a approximation of growth for big n:s, you need to know "twice what" and "four times what" and "what's a big n" to have any use from it.

For the example with multiplication of 2 n-digit numbers:

https://en.wikipedia.org/wiki/Karatsuba_algorithm It's O(n^1.585). Need about 1000 digits before it's actually faster than "School book" O(n^2) multiplication.

https://en.wikipedia.org/wiki/Sch%C3%B6nhage%E2%80%93Strasse... It's O(n log n log log n). In practice faster than Karatsuba at 10,000 to 40,000 decimal digits.

Then there is the newest and hottest O(n log n) version, that's not practical at all.

Sure, there are some cases where O() can be misleading, but those are the exceptions. You can work in software for a long time without being affected significantly by them, whereas using O(n^2) when you could use O(n) can make your code 1,000 times slower than usable, and it does this frequently.

I have fixed hundreds of O(n^2) algorithms over the years, made them O(n) or O(n log(n)), and made the product dramatically better. The fact that an O(n) algorithm can sometimes be slower than an O(n^2) feels more like pedantry than useful information in this context.