You cannot write $O(f(n)) = n^3$ or $O(f(n)) = \log n$ since this is a "syntax error". Big O notation is not symmetric, and can only be used on the right-hand side, or in a chain of inequalities $f(n) = O(g(n)) = O(h(n))$ (formally, this means $f(n) \in O(g(n)) \subseteq O(h(n))$).
Your function $f(n)$ satisfies $f(n) = \Theta(n^2)$, as well as $f(n) = \Theta(n^2 + 2n + \log n)$, $f(n) = \Theta(n^2 + n)$, $f(n) = \Theta(17n^2 + n/\log n + 5\log\log n)$, and many more. Regarding big O and big Omega, there are even more options, for example $f(n) = O(n!^2)$ and $f(n) = \Omega(1/n)$.
That said, when stating big Theta estimates, we are usually aiming for the simplest possible expression on the right, so in your case we are most satisfied with $f(n) = \Theta(n^2)$. 675
With big O and big Omega, we are looking for both simplicity and tightness, which is why we would be most satisfied with $f(n) = O(n^2)$ and $f(n) = \Omega(n^2)$ ($f(n) = \Omega(\log n)$, for example, is simple but not tight).
The notation $f(n) = \Theta(g(n))$ means that for some constants $c,C>0$, $$ c \leq \frac{f(n)}{g(n)} \leq C. $$ In other words, $f(n),g(n)$ are the same up to a (multiplicative) constant. The notation $f(n) = O(g(n))$ corresponds to just the upper bound $f(n) \leq Cg(n)$, and $f(n) = \Omega(g(n))$ corresponds to just the lower bound $f(n) \geq cg(n)$.
If you think of $f(n) = \Theta(g(n))$ as $f(n) = g(n)$ (up to constants), then $f(n) = O(g(n))$ is like $f(n) \leq g(n)$, and $f(n) = \Omega(g(n))$ is like $f(n) \geq g(n)$.
The notation $f(n) = o(g(n))$ just means that $f(n)/g(n) \to 0$, and $f(n) = \omega(g(n))$ just means that $f(n)/g(n) \to \infty$, so that $f(n) = o(g(n))$ is the same as $g(n) = \omega(f(n))$.
(Caveat: the discussion above assumes all functions are positive.) 804