Understanding the reduction of REGULARTM from ATM
Let's look just at $M_2$ for the moment. If $M$ accepts $w$, then $M$ will accept strings of the form $0^n1^n$ in step (1) and everything else in step (2). In other words

If $M$ accepts $w$, then the language $L(M_2)=\Sigma^*$, which is of course regular.

Now if $M$ doesn't accept $w$, then $M_2$ will accept all and only strings of the form $0^n1^n$, so $L(M_2) = \{0^n1^n\mid n\ge 0\}$ and this language is certainly not regular. 

The important thing here is that we don't have to run $M$ on $w$ to decide its behavior: the behavior is determined solely by whether $M$ accepts $w$ or not. Since $R$ was assumed to be a decider, it doesn't have to run $M$ to determine whether the language of $M_2$ is regular or not; because $R$ is a decider it is by definition assumed to be able to tell that.