This question is related to this one.
Probably it's obvious but could you tell me what is the grading on the graded direct product?
I was thinking about $^*\Pi M^i=\oplus_j(\Pi_i M^i_j)$ where $M^i_j$ is the degree $j$-part of $M^i$, but I couldn't prove it and I'm not sure I'm correct, especially because if $R=M^i=k[x]$, then $(1,x,x^2,x^3,\ldots)$ seems to belong to the graded direct product but I don't think it belongs to $\oplus_j(\Pi_i M^i_j)$.
It seems you've misunderstood the definition of $^*\Pi M_i$ in Bruns and Herzog. When write $^*\Pi M_i$ is "the submodule of $\Pi M_i$ generated by the sequences $(x_i)$ with $x_i$ homogeneous" you should add "of degree $n$, $n\in\mathbb Z$", that is, all $x_i$ have the same degree, and thus their definition coincides (how else?) with $\oplus_j(\Pi_i M^i_j)$.