Could someone give me a simple explanation why an oracle machine that can solve the halting problem for standard Turing machines, is however unable to solve the halting problem for other such oracle machines?
I have tried searching for the answer online but could not find any that really answered the question.
Remember that an oracle machine isn't really a "complete object" - basically anything interesting we might ask of it depends on what oracle we feed it. For example, whether an oracle machine $\Phi_e^-(e)$ halts or not depends in general on what oracle it's working with.
So let me rephrase the fact you're starting with:
There is an oracle $X$ and an oracle machine $\Phi_e^-$ such that $\Phi_e^X$ (= $\Phi_e^-$ with oracle $X$) computes the halting problem.
Now every specific oracle has a corresponding halting problem: namely, given an oracle $X$ we let $$X':=\{e: \Phi_e^X(e)\mbox{ halts}\}.$$ The usual proof that the halting problem is not computable translates immediately to prove that $X'$ is not $X$-computable - that is, for every oracle $X$, there is no oracle machine $\Phi_e^X$ such that $\Phi_e^X$ computes $X'$.
Since $X'$ depends heavily on $X$, there is no "halting problem for oracle machines" - rather, each oracle determines a different "relativized halting problem," and the more complicated we make the oracle the more complicated this becomes, with the result that we never "catch our tail."
EDIT: here's how to "relativize" the unsolvability of the halting problem:
Fix an oracle $X$. Suppose $c$ "solves the $X$-halting problem" - that is, for each $n$ we have $\Phi_c^X(n)=1$ iff $n=X'$. As in the non-oracle case, there is$^*$ an oracle machine $\Phi_e^-$ such that for all $n$, we have $\Phi_e^X(n)\downarrow$ iff $\Phi_c^X(n)\downarrow=0$. Then $\Phi_c^X(e)=0$ iff $e\in X'$, contradicting the assumption on $c$.
$^*$This uses the relativized version of the existence of a universal machine, which is proved analogously as for non-oracle machines. Note, incidentally, that $e$ is independent from $X$: a single $e$ does the job for every oracle.
The proof that a Turing machine with an oracle for $X$ can't solve the halting problem for Turing machines with an oracle for $X$ is identical to the proof that an ordinary Turing machine can't solve the halting problem for ordinary Turing machines.
