There might have been confusion on the meaning of "return the opposite of what it returns" when you run $M$ on $w$.
When $M$ runs forever, nothing can be returned by $M$. Then that rule does not make sense! Not only there is no way to specify "the opposite of" nothing, but also it is simply impossible to return, i.e., to halt.
By the way, note that there is huge difference between "halt and return nothing" and running forever. In the case of running forever, a running machine does not return anything, either. Note that "it returns nothing" might be ambiguous.
The right specification is to stipulate "return the opposite of what is returns when it halts". The condition "when it halts" should be present, although it might be skipped in sloppy writing or in well-understood context.
The better choice could be sticking to "accept", "reject", "halt" and "loop" and abandoning the word "return" when we are talking about decision problems.
Since the question involves Turing machines that can reject, let us define a Turing machine as an 8-tuple $\langle Q,\Gamma ,b,\Sigma ,\delta ,q_{0},A, R\rangle$, where $Q,\Gamma ,b,\Sigma ,\delta ,q_{0}$ are as usual and $A$ and $R$ are the set of accepting and rejecting states respectively. A Turing machines halts means it either accepts or rejects and vice versa.
Here is a clearer description of the wanted reduction.
Given $\langle M,w\rangle$, switch the set of accepting states and the set of rejecting states in $M$ to obtain a new Turing machine $M'$. More specifically, $M$ is the same as $M'$ except that each accepting state in $M$ will be defined as a rejecting state in $M'$ and each rejecting state in $M$ will be defined as an accepting state in $M'$. More formally, if $M=\langle Q,\Gamma ,b,\Sigma ,\delta ,q_{0},A, R\rangle$, then $M'=\langle Q,\Gamma ,b,\Sigma ,\delta ,q_{0},R, A\rangle$.
Let $f:\Sigma^*\to\Sigma^*$ be defined by $f(u)=\langle M', w\rangle$ if $u=\langle M,w\rangle$ and $f(u)=u$ otherwise.
We can check routinely that the map $f$ is a one-to-one Turing reduction from $REJECT$ to $ACCEPT$ as well as a one-to-one Turing reduction from $ACCEPT$ to $REJECT$.
