How does the split mask countermeasure work on AES?
Given some intermediate data $x$ as two shares $x=x_1\oplus x_2$ take some fresh random $r$ to calculate new shares $x_1' = ((x_1\oplus r)\oplus x_2)\oplus(n\oplus r)$ [parenthesis indicating the order of evaluation] and $x_2' = n$. Now you can use $x_1'$ ($=x\oplus n$) as input for both tables.

The answer to "So how should it be computed?" is not at all. It is not needed. I think this is just a warning that it would be a problem to have (2) as intermediate value, because $m$ is kept constant for many plaintexts. Strangely enough, the inventors of the split mask countermeasures didn't realize that keeping $n$ constant for many plaintexts poses the same risks. A very short look at the paper you linked to, gave me the impression that it will expose this weakness in later chapters.