Qfyilyi

I totally understand what big $O$ notation means. My issue is when we say $T(n)=O(f(n))$ , where $T(n)$ is running time of an algorithm on input of size $n$.

I understand semantics of it. But $T(n)$ and $O(f(n))$ are two different things.

$T(n)$ is an exact number, But $O(f(n))$ is not a function that spits out a number, so technically we can't say $T(n)$ equals $O(f(n))$, if one asks you what's the value of $O(f(n))$, what would be your answer? There is no answer.

Strictly speaking, $O(f(n))$ is a set of functions. So the value of $O(f(n))$ is simply the set of all functions that grow asymptotically not faster than $f(n)$. The notation $T(n) = O(f(n))$ is just a simplified way to write that $T(n) in O(f(n))$.

Note that this also clarifies some issues of the $O$ notation as it is normally used.
For example, $n^2 in O(n^3)$ but $n^3 notin O(n^2)$. So you can write $n^2 = O(n^3)$, but you cannot write $n^3 = O(n^2)$.

Just as a small contribution ... In The Algorithm Design Manual book, you can find a paragraph about this issue:

The Big $O$ notation provides for a rough notion of equality when comparing
functions. It is somewhat jarring to see an expression like $n^2 = O(n^3)$, but its
meaning can always be resolved by going back to the definitions in terms of upper
and lower bounds. It is perhaps most instructive to read the " = " here as meaning
"one of the functions that are". Clearly, $n^2$ is one of functions that are $O(n^3)$.

This is in line with Vincenzo's answer: you should simply interpret $O(f(n))$ as a set of functions and the = symbol as a set membership symbol $in$.

$O$ is a function
$$begin{align}
O : (mathbb{N}to mathbb{R}) &to mathbf{P}(mathbb{N}to mathbb{R})
\ f &mapsto O(f)
end{align}$$
i.e. it accepts a function $f$ and yields a set of functions that share the asymptotic behaviour of $f$. And strictly speaking the correct notation is thus
$$
(n mapsto T(n)) in O(nmapsto f(n))
$$
or short
$$
T in O(f)
$$
but it's customary in maths, science and CS to just use a variable somewhere in the expression to denote that you're considering functions of the argument $n$ on both sides. So $T(n) in O(f(n))$ is quite fine as well. $T(n) = O(f(n))$ is pretty much wrong though, as you suspected.