Base change of $H^n(\Gamma, \mathrm{Sym}^{k}(R^2))$ - A small step in Eichler-Shimura I'm currently learning Eichler-Shimura mapping and found the note by Gabor Wiese is quite helpful. Yet I have come up with a quite detailed question.
Let $R$ be a ring (all rings in this post are commutative with identity). Let $V_{k}(R) := R[X,Y]_{k} = \mathrm{Sym}^{k}(R^2)$, where $R[X,Y]_{k}$ is the abelian group of all homogeneous polynomials in two variables $X$ and $Y$ of degree $k$. (Defined on page 12 of the note linked above)
Let $\Gamma \leq \mathrm{SL}_2(\mathbb{Z})$ be a subgroup of finite index. We can define the action of $\Gamma$ on $V_{k}(R)$ as
$$
    \left( \begin{pmatrix}
    a & b \\ c & d
    \end{pmatrix} \cdot f \right) (X,Y) := f \left( (X,Y) \begin{pmatrix}
    a & b \\ c & d
    \end{pmatrix} \right) = f(aX+cY, bX+dY), 
    $$
for all $\begin{pmatrix}
    a & b \\ c & d
    \end{pmatrix} \in \Gamma$ and $f \in V_k(R).$
We can check that then $V_{k}(R)$ is a $\Gamma$-module, and hence we can consider the group cohomology $H^{\bullet}(\Gamma, V_{k}(R))$.
Now let $\phi: R \rightarrow S$ be a ring homomorphism (or in other words, $S$ is an $R$-algebra), then my question is: does the following holds?
$$
H^n(\Gamma, V_k(R)) \otimes_{R} S \cong H^n(\Gamma, V_k(S))   \quad \quad (\star_n).
$$
I have read quite a lot of notes on these things yet it seems that everyone takes this for granted (at least for $n=1$ case, i.e. $(\star_1)$)? Does this hold for general $n$? Yet I have tried on proving this but failed. Sorry for being stupid and I'm not quite familiar with the base change stuff here and hoping that this can be a start for me to get familiar with such things.
(It seems that some of such base change property is obvious and some of them are quite deep. Unfortunately, I cannot distinguish the two now. :( )
Any further references are welcome as well.
Edit: At least I hope that $(\star_{1})$ holds in the two following cases:

*

*when "$R \rightarrow S$" is $\mathbb{Z} \rightarrow \mathcal{O}$;

*when "$R \rightarrow S$" is $\mathcal{O} \rightarrow \overline{\mathbb{Q}_p}$,
where $E  | \mathbb{Q}_p$ is a finite extension and $\mathcal{O}$ is its ring of integers.

Edit: My motivation for this question comes from the Page 3 of the note Hecke algebra valued Galois representations for $\mathrm{GL}_2(\mathbb{Q})$, where the above two cases are claimed.
Sorry for the edit and thank you all for noticing this post!
EDIT I have deleted my further edit on Kunneth spectral sequence and put it as an answer below.
And still, I'm wondering if there are any solutions without knowing the knowledges on spectral sequence?
Thank you all again for kind comments and sorry for being so lengthy and maybe talkative.
 A: EDIT: There're something WRONG as Professor David Loeffler pointed out in the comment. I shall correct this these days! Sorry for that!
Though it seems that few people are interested in this problem, I shall post an answer to show what I have got. What follows is mainly filling up the details in Professor David Loeffler's great comment.
The main tool is the Kunneth spectral sequence, as the following theorem provides:

Theorem: (Weibel's Intro. to homological algebra, Theorem 5.6.4) Let $P$ be a bounded below complex of flat $R$-modules and $M$ an $R$-module. Then there is a boundedly converging right half-plane spectral sequence
$$
E_{pq}^2 = \mathrm{Tor}_p^R (H_q(P), M) \Rightarrow H_{p+q}(P \otimes_{R} M).
$$

Turn this into a cohomological version, we have
$$
E^{pq}_2 = \mathrm{Tor}^p_R (H^q(P), M) \Rightarrow H^{p+q}(P \otimes_{R} M).
$$
Now put $M=S$ here, and $P$ the cochain complex $C^{\bullet}(\Gamma, V_{k}(R))$. Then the right hand side of "$\Rightarrow$" turns into $C^{\bullet}(\Gamma, V_{k}(R)) \otimes_{R} S$. So to use this spectral sequece, we need to check that
$$ C^{\bullet}(\Gamma, V_{k}(R)) \otimes_{R} S = C^{\bullet}(\Gamma, V_{k}(S)) .$$
It seems that this is indeed a routine check.
Hence, we obtain the spectral sequence
$$
E^{pq}_2 = \mathrm{Tor}^p_R (H^{q}(\Gamma, V_k(R)), S) \Rightarrow H^{p+q}(V_k(S)).
$$
Using the long exact sequence deduced from this (see Neukirch's "Cohomology of number fields", Lemma 2.1.3), we have
$$
0 \rightarrow \mathrm{Tor}^1_R (H^{0}(\Gamma, V_k(R)), S)) \rightarrow H^1(\Gamma, V_k(S)) \rightarrow \mathrm{Tor}^0_R (H^{1}(\Gamma, V_k(R)), S) \rightarrow \mathrm{Tor}^2_R (H^{0}(\Gamma, V_k(R)), S) \rightarrow \cdots .
$$
Note that $\mathrm{Tor}^0_R (H^{1}(\Gamma, V_k), S)) = H^1(\Gamma, V_k(R)) \otimes_{R} S$, we obtain
$$
0 \rightarrow \mathrm{Tor}^1_R (H^{0}(\Gamma, V_k(R)), S) \rightarrow H^1(\Gamma, V_k(S)) \overset{(\star)_1} \rightarrow H^1(\Gamma, V_k(R)) \otimes_{R} S \rightarrow \mathrm{Tor}^2_R (H^{0}(\Gamma, V_k(R)), S) \rightarrow \cdots .
$$
Hence to make $(\star)_1$ an isomorphism, it suffices to require the vanishing of
$ \mathrm{Tor}^1_R (H^{0}(\Gamma, V_k(R)), S)$ and $\mathrm{Tor}^2_R (H^{0}(\Gamma, V_k(R)), S)$, which means we need at least one condition below holds:

*

*$S$ is flat over $R$;

*$H^{0}(\Gamma, V_k(R))$ is flat over $R$.

This is what Professor David Loeffler said in the comment.
However, I haven't found the way to solve for $(\star)_n$ in general, since the term "$\mathrm{Tor}_{R}^0$" only appears on the first terms of the induced long exact sequence. (Yet the $n=1$ case is enough to me up to now.)
As a new learner, I'm not quite sure that the above solution is correct. And I'm still hoping to get the general $n$ case, as well as a solution without the spectral sequence.
Thank you all for paying attention! Sincerely sorry for possible mistakes!
