let's consider that we already have constructed heap array.
so from this, when we do heap sort, the number of elements that have to be sorted decreased. I mean heap decrease.(which also means heap tree level decreases -> downheap operation level decreases
but nlogn means each n nodes has logn operation. But not all nodes do logn(original down heap operation level) operation because of heap tree level decrease.
$n \log n$ means each $n$ nodes has $\log n$ operation.
No, it does not. That's like saying:
\$100 means I sold 100 drinks for \$1 each
Obviously, you could have sold any number of things as long as they add up to \$100.
With algorithm costs it's the same. A sequence of operations taking "time" $\Theta(n \log n)$ means that the "time" costs of all the operations add up to something that behaves roughly like $n \log n$.
Once Landau terms enter the pickture, things get muddled. It's tempting to tread $\sum_{i=1}^n O(\log i)$ in the same way as $\sum_{i=1}^n \log i$, but the former sum doesn't really make sense. I recommend you read our reference questions on algorithm analysis and asymptotics to see how one can arrive at the final result here.
