Commutativity of Tor and restriction Let $R=S[\mathbf{x},\mathbf{y}]$ be a standard bigraded polynomial ring in the sets of variables $\mathbf{x}=\{x_1,...,x_m\}$ and $\mathbf{y}=\{y_1,...,y_m\}$ (i.e., $\deg(x_i)=(1,0)$ and $\deg(y_j)=(0,1)$. Consider the standard polynomial ring $T=S[\mathbf{x}]$ (that can be viewed as a bigraded ring contained in $R$) and fix an integer $a$.
Now, given a bigraded $R$-module $M$ we define the graded $T$-module $M_{*,a}=\bigoplus_{r\in\mathbb Z}(M_{*,a})_r$ with $(M_{*,a})_r=M_{r,a}$. I wonder if there is a functorial isomorphism $$\DeclareMathOperator{\Tor}{Tor} \Tor_p^R(M,R/(\mathbf{y}))_{*,a}\simeq\Tor_p^T(M_{*,a},S)$$ for any $p\geq0$.
More generally, if $R=S[\mathbf{x}_1,...,\mathbf{x}_n]$ is standard $\mathbb Z^n$-graded, $T=S[\mathbf{x}_{i_1},...,\mathbf{x}_{i_s}]$ and $\nu\in\mathbb{Z}^{n-s}$, is there functorial isomorphism $$\Tor^R_p(M,R/(\mathbf{x}_{i_1},...,\mathbf{x}_{i_s}))_{*,\mu}\simeq\Tor^T_p(M_{*,\mu},S)$$ for any graded $R$-module $M$ and $p\geq0$?
 A: Yes, this is correct. One way to proceed is as follows.
Let $j:T\to R$ denote the inclusion of $T$ into $R$. If $M$ is an $R$-module, then what you are observing is that the $T$-module $j^*(M) = M$ is a direct sum of $T$-invariant subspaces given by $M^a = \bigoplus_{r\in \mathbb Z} M_{r,a}$.
Moreover, $R$ is a free $T$-module. Now consider the change of rings spectral sequence for the inclusion $j$, which looks like for a $T$-module $N$ and an $R$-module $M$, as follows:
$$E_{p,q}^2 = \mathrm{Tor}_p^R(\mathrm{Tor}_q^T(N,R),M)\Longrightarrow_p \mathrm{Tor}_{p+q}^T(N,j^*(M)).$$
In your case, let us choose $N=S$ the ground ring , and let $M$ be some bigheaded $R$-module. Because $R$ is $T$-free, it follows that
$$\mathrm{Tor}_q^T(S,R)= \begin{cases} 0 & \text{ if $q\neq 0$ }\\ 
   S\otimes_T R = R/(x)& \text{ else.}
\end{cases} $$
In particular, $E^2_{p,q}=0$ is zero unless $q=0$, so the sequence degenerates at this page, and
$$E^2_{p,0} = \mathrm{Tor}_p^R(R/(x),M).$$
This is an $R/(x)\simeq T$-module, so its has its "$a$-decomposition". At each fixed $a\in\mathbb Z$, it follows that the spectral sequence converges to the $T$-module
$$\mathrm{Tor}_p^T(S,M^a)$$
which we just proved is equal to $\mathrm{Tor}_p^R(R/(x),M)^a$. Your general claim should follow mutatis mutandis.
One way to summarise the statement is: if $T\to R$ is any ring morphism such that $R$ is $T$-free, then restriction of $R$-modules to $T$-modules induces an isomorphism on $\mathrm{Tor}$ given by
$$ 
\mathrm{Tor}_p^R(M\otimes_T R,N) \longrightarrow \mathrm {Tor}_p^T(M,j^*(N)).
$$
In fact, you needn't really use a spectral sequence: what this shows is that any $R$-free resolution of some $M$, under pullback, is a $T$-free resolution of $j^*(M)$, and you can proceed from there.
