Can the higher-order oracle Turing machines simulate the lower-order machines so that the current oracle does not contradict the simulated oracle?
The halting problem $\mathrm{Halt}_O$ for Turing machine with oracle access to $O$ is the language consisting of all oracle Turing machines $M$ that halt on the empty input when the oracle calls are interpreted as membership in $O$. A Turing machine with access to $\mathrm{Halt}_O$ can compute the busy beaver number for oracle Turing machines with oracle access to $O$.

Now to your other question. Let $O_0 := \emptyset$, and $O_{i+1} = \mathrm{Halt}_{O_i}$ (so $O_i = 0^{(i)}$ using the usual terminology). Let us show that given oracle access to $O_{i+1}$ you can simulate oracle access to $O_i$. Consider the following program:


  If $x \in O_i$ then halt, else run an infinite loop.


Using oracle access to $O_{i+1} = \mathrm{Halt}_{O_i}$ you can determine whether this program halts, which happens if and only if $x \in O_i$.

Applying this construction inductively, you can simulate access to $O_i$ given access to $O_j$ for all $j \geq i$. This shows that having an oracle for $O_j$ is indeed stronger than having an oracle for $O_i$ (when $i &lt; j$).