Let me start of with a recommendation: treat Landau notation just as you (should) treat rounding: round rarely, round late. If you know something more precise than $O(.)$, use it until you are done with all calculations, and Landauify at the end.
As for the question, let's dig through this abuse of notation¹. How would we interpret something like $h \in O(f + O(g))$? We should replace $O$ with its definition from the inside out. So, we get
$\qquad \displaystyle \exists g' \in O(g).\, h \in O(f + g')$
$\qquad \displaystyle \exists g' \in O(g).\,\exists d>0.\, \forall n.\, h(n) \leq d(f(n) + g'(n))$
which is equivalent to
$\qquad \displaystyle \exists c > 0.\,\exists d>0.\, \forall n.\, h(n) \leq d(f(n) + cg(n))$.
As certainly² $d(f(n) + cg(n)) \leq cd(f(n) + g(n))$, we see that this is equivalent to $h \in O(f + g)$; the loss of precision is ignored by $O$ anyway.
What about other combinations, say $h \in O(f + \Omega(g))$? If we try the same here, we get
$\qquad \displaystyle \exists g' \in \Omega(g).\, h \in O(f + g')$.
But this is a tautology: $h$ is certainly bounded above by something arbitrarily large. So, combining upper and lower bounds in this fashion is not meaningful.
- $O(.)$ and the other Landau symbols map functions to function classes. Feeding it a function class is not immediately meaningful.
- At least if we consider only positive functions, which we can safely assume when talking about runtimes. I'm not sure this works in general.