Proving big-theta complexity with constants in $f(n)$
You can find the definitions of big O notations on Wikipedia. They agree with your definition of big $\Theta$.
As for your actual question, the answer is actually quite subtle. Sometimes it is possible to find constants $c_1,c_2,n_0$ which work for all values of the parameters. For example, in your case, all $n \geq 1$ satisfy
$$
1 \cdot n^{\max(a,b)} \leq n^a + n^b \leq 2 \cdot n^{\max(a,b)},
$$
and so you can say that $n^a + n^b = \Theta(n^{\max(a,b)})$.
(As a side note, you can also say that $n^a + n^b = \Theta(n^a + n^b)$, or that $n^a + n^b = \Theta(n^a + 2n^b)$; but usually we aim at the simplest possible expression, or an expression of a particular form, say $n^\alpha \log^\beta n$.)
In other cases we are not so lucky. For example, $\log^k n = O(n)$ for all $k$, but the relevant constants depend on $k$. Sometimes this makes no difference, but in other cases it is important to know that the constants depend on $k$. When it does matter, we sometimes use the notation $O_k(n)$ to stress that the hidden constants depend on $k$. This is not completely standard, but pretty common. When the hidden constants are independent of some parameter, we can explicitly say so – $n^a + n^b = \Theta(n^{\max(a,b)})$, where the hidden constants are independent of $a,b$.
In your case, since you are only interested in an isolated estimate, you don't care about the dependence of the hidden constants on the parameters. So for you, $\log^k n = O(n)$ holds, even though the hidden constants depend on $k$, since for any particular $k$ this statement holds. This is the usage in the master theorem, for example – the constants there depend on the parameters, but usually it makes no difference, because the parameters are fixed in any given application, in the vast majority of circumstances.