If you consider $S$ as you propose, that is $\bigoplus\limits_{n_i,n_j=0}T_{\underline{n}}$, then $T$ is a bi-graded ring over $S$: $$S_{p,q}=\bigoplus\limits_{n_i=p,n_j=q}T_{\underline{n}}.$$ By the choice the author makes, they just ensure to isolate the grading in one direction, thus making it a $\mathbb{Z}$-graded ring.

If you consider $S$ as you propose, that is $\bigoplus\limits_{n_i,n_j=0}T_{\underline{n}}$, then $T$ is a bi-graded ring over $S$: $$T_{p,q}=\bigoplus\limits_{n_i=p,n_j=q}T_{\underline{n}}.$$ By the choice the author makes, they just ensure to isolate the grading in one direction, thus making it a $\mathbb{Z}$-graded ring.