If $M_*$ and $N_*$ are graded modules over the *graded* ring $R_*$, what is the definition of $M_* \otimes_{R_*} N_*$?

Usually the tensor product of two graded $R_\bullet$-modules $M_\bullet, N_\bullet$ is defined as having $n$-th component $$\left(M_{\bullet}\otimes_{R_\bullet} N_\bullet\right)_n\ :=\ \left(\bigoplus\limits_{p+q=n} M_p\otimes_{\mathbb Z} N_q\right)\left/\left(m r\otimes n - m\otimes r n\ |\ a+b+c=n, m\in M_a, r\in R_b, n\in N_c\right)\right..$$ In particular, this is a quotient of $\bigoplus\limits_{p+q=n} M_p\otimes_{R_0} N_q$, but usually much smaller.

For example, take $R_\bullet=M_\bullet=N_\bullet:={\mathbb k}[x]$ for some commutative ring $\mathbb k$, with $\text{deg}(x)=1$. Then $$\bigoplus\limits_{p+q=n} M_p\otimes_{R_0} N_q={\mathbb k}[x,y]_n,\quad\text{ while }\quad(M_\bullet\otimes_{R_\bullet} N_\bullet)_n=R_n={\mathbb k}[x]_n.$$

Just a comment -- an indirect but intuitive way to remember this is the following:

First of all, as a module $M_* \otimes_{R_*} N_*$ is just the ordinary tensor product of $R_*$-modules. This is the only reasonable construction (we can always apply the operation of "forgetting the grading", which reduces the construction to this ordinary tensor product).

So the module structure is unique. That leaves the grading: observe that $M_* \otimes_{R_*} N_*$ is generated by homogeneous elements of the form $m \otimes n$. Clearly, the degree of such an element is $\deg(m) + \deg(n)$.

Moreover, the defining relations of $R_*$-bilinearity respect this rule -- multiplying by scalars raises the degree in the `intuitive' way -- which means it's well-defined to grade elements of $M_* \otimes_{R_*} N_*$ by (a) finding any way to represent them as sums of simple tensors, then (b) using the above rule.

Following this reasoning a few more steps leads to the nice computation done by Hanno.