Why is TIME(n log (log n)) \ TIME(n) = ∅?
The quick explanation is that 

$TIME[o(n\log n)]\subseteq REG\subseteq TIME[n]\subseteq TIME[o(n\log n)]$, and therefore $TIME[o(n\log n)]=REG$. 

Similarly:

$TIME[n\log\log n]\subseteq TIME[o(n\log n)]\subseteq REG\subseteq TIME[n]\subseteq TIME[n\log \log n]$, so $TIME[n\log\log n]=REG$

But I think this is not the point you are missing.

You say that $TIME[n\log \log n]$ is regular. This is not exact. When we say that something 
 is regular, we mean that it is a language $L\subseteq \Sigma^*$, which is regular (i.e. can be recognized by a DFA).

The class REG is not a language, but a set of languages. That is, $REG\subseteq 2^{\Sigma^*}$. Similarly, $TIME[f(n)]\subseteq 2^{\Sigma^*}$ for every function $f$. These are all classes of languages.

Since we have that $TIME[n\log\log n]\subseteq TIME[o(n\log n)]\subseteq REG$, then 
$REG\setminus TIME[n\log\log n]=\emptyset$. This follows from the simple property that if $A\subseteq B$, then $A\setminus B=\emptyset$.