Acceptance conditions when translating LTL to Büchi automaton?
I will start with the last question and then come to the question at the top of your post.


The construction assumes that the LTL formula is given in negation normal form, so that no temporal operator occurs in negated form (as you wrote).
So $\neg (p_1 U (p_2 U p_3))$ gets translated to $\neg p_1 R (\neg p_2 R \neg p_3)$, and because in the latter formula, there is no Until subformula, we have 0 acceptance pairs in the Generalized Buchi automaton.
Now, a generalized Buchi automaton accepts all words for which there exists an infinite run on which all acceptance sets are visited infinitely often. If there are none, then this condition is vacuously true on any infinite run, which still has to exist. For example the word $(\{p_1,p_2\}^\omega)$ does not satisfy the LTL formula, and the corresponding generalized Buchi automaton has not infinite run for the word. If you model check with this automaton, every reachable cycle represents a case in which the LTL property holds. If you translate the generalized Buchi automaton to a Buchi automaton, then the standard construction makes every state in the latter accepting.
As far as the subformulas are concerned, you have to take the definition quite literally. For every Until-subformula in the NNF LTL property, you have an acceptance set. For instance, for $(a\,\mathcal{U}\,b)\,\mathcal{U}\,c$, you will have acceptance sets for $(a\,\mathcal{U}\,b)\,\mathcal{U}\,c$ and $(a\,\mathcal{U}\,b)$. For the All states in the generalized Buchi automaton containing $b$ or not containing $(a\,\mathcal{U}\,b)$ as a disjunct are accepting for  the latter acceptance set. All states containing $c$ as a disjunct or not containing $(a\,\mathcal{U}\,b)\,\mathcal{U}\,c$ as a disjunct are labeled as contained in the acceptance set for $(a\,\mathcal{U}\,b)\,\mathcal{U}\,c$.
The example $((\neg A)R(\neg B))\vee(\top U C)$ that you mentioned is actually not problematic. Depending on how you write down the algorithm, either (1) no state will ever be labeled by $((\neg A)R(\neg B))\vee(\top U C)$, but only by $((\neg A)R(\neg B))$ or $(\top U C)$ (and you will have two initial states), or (2) all states labeled by $((\neg A)R(\neg B))\vee(\top U C)$ need to also be labeled by either $((\neg A)R(\neg B))$ or $(\top U C)$, and for determining in which acceptance set a state is, you only look at the whole subformulas, so a state labeled by $((\neg A)R(\neg B))$ would be accepting for all acceptance sets, while one also labeled by $(\top U C)$ would not be labeled by the acceptance set for $(\top U C)$ (unless it is also labeled by $C$).