I am confused about how this Tor is defined.
Suppose $R$ is a graded ring, $M,N$ graded modules over $R$. What is $\operatorname{Tor}_{st}^R(M,N)$?
I am confused about the subscripts. I realize there is a lot of grading, so a definition that clarifies the grading and the subscripts would be helpful.
My favorite explanation of the grading on this object is in the preliminaries to (the published version of) Paul Baum's dissertation, linked here.
One first forms a (graded) projective resolution $P_\bullet = (P_s)$ of $M$, with degree-0 maps of graded modules $f_s\colon P_{s} \to P_{s-1}$, then removes the segment $P_0 \overset{f_0}\to M$, replacing it with $P_0 \to 0$.
Then one tensors the whole thing with $N$ to get a complex $P_\bullet \otimes_R N = (P_s \otimes_R N)$, with maps $f_s \otimes \mathrm{id}_N$. One grades the tensor product by total degree $t$, so $$(P_s \otimes_R N)^{(t)} = \bigoplus_{a+b = t} \big(P_s^{(a)} \otimes_R N^{(b)}\big).$$ The maps $f_s \otimes \mathrm{id}_N$ in this complex take bidegree $(s,t)$ to bidegree $(s-1,t)$. For fixed $t$, the Tor with subscript $(s,t)$ is the $s$th homology of the complex $(P_\bullet \otimes_R N)^{(t)}$; that is, $$\mathrm{Tor}_{s,t}^R(M,N) = \frac{\ker\!\big((f_s \otimes \mathrm{id}_N)^{(t)}\colon \ (P_s \otimes_R N)^{(t)} \to (P_{s-1} \otimes_R N)^{(t)} \big)} {\mathrm{im}\big((f_{s+1} \otimes \mathrm{id}_N)^{(t)}\colon \ (P_{s+1} \otimes_R N)^{(t)} \to (P_s \otimes_R N)^{(t)}\big)}. $$
If you want to forget the grading on $R$, $M$, and $N$, then, summing along $t$, you can, and you recover $$\mathrm{Tor}^R_s(M,N) = \mathrm{Tor}^R_{s,\bullet}(M,N) = \bigoplus_{t\geq 0} \mathrm{Tor}^R_{s,t}(M,N),$$ the regular, ungraded $s$th Tor functor. As in the ungraded case, the results turn out to be independent of the resolution chosen.
It's also common to cohomologically grade by swapping $-s$ for $s$ everywhere. One then writes $$\mathrm{Tor}^{-s,t}_R(M,N) = \mathrm{Tor}_{s,t}^R(M,N).$$