What is true is that $\rm Tor$ commutes with filtered colimits. We already know that $M\otimes -$ commutes with colimits, now take a right $R$-module $M$ and a system $(N_i,\psi_{ji})$ of left $R$-modules over some filtered set $I$. First, we can obtain a short exact sequence of filtered systems
$$0\to (K_i,\rho_{ji})\to (P_i,\tilde\psi_{ji})\to (N_i,\psi_{ji})\to 0$$
by covering $N_i$ by $P_i=R^{(N_i)}$ by $e_n\to n$ and defining $\tilde \psi_{ji}(e_n)=e_{\psi_{ji}n}$, the condition $\psi_{kj}\psi_{ji}=\psi_{ki}$ is immediately inherited by the $\tilde \psi_{ji}$, similarly for the induced morphisms $\rho_{ji}:K_i\to K_j$. Since $I$ is filtered, $\rm colim$ is exact giving an exact sequence
$$0\to {\rm colim}\; K_i \to {\rm colim}\; P_i \to {\rm colim}\;N_i\to 0$$
Call the terms $K,P,N$ for brevity. Since each $P_i$ is flat and $I$ is filtered, $P$ is flat. We may then use the long exact sequence for $\rm Tor$ and obtain a diagram
$\require{AMScd}$
$$\begin{CD}
\\
{\rm Tor_1}\;(M,P) @>>>
{\rm Tor_1}\;(M,N) @>>>
M\otimes K @>>>
M\otimes P
\\
{}&{}& {} &{}& @VVV @VVV
\\
{\rm colim}\;{\rm Tor_1}\;(M,P_i)@>>>
{\rm colim}\;{\rm Tor_1}\;(M,N_i)@>>>
{\rm colim}\; M\otimes K_i@>>>
{\rm colim}\; M\otimes P_i' \\
\end{CD}$$
The first two columns vanish since $P_i,P$ are flat, and the last two columns are connected by natural isomorphisms that give a commutative diagram. You can check that if you have an incomplete commutative diagram with exact rows
$$\begin{CD}
\\
0@>>>
A @>>>
B @>>>
C \\
{}&{}& {} &{}& @VVV @VVV
\\
0 @>>>
A'
@>>>
B' @>>>
C' \\
\end{CD}$$
you may always complete it, and the morphism introduced is an isomorphism if both vertical arrows are. This gives the isomorphism for $n=1$. The case $n\geqslant 1$ is handled by dimension shifting. Indeed, we get a diagram
$$\begin{CD}
\\
0@>>>
{\rm Tor}_2(M,{\rm colim}\;N_i) @>\partial >>
{\rm Tor}_1(M,{\rm colim}\;K_i) @>>>
0 \\
{}&{}& {} &{}& @VVV
\\
0 @>>>
{\rm colim}\;{\rm Tor}_2(M,N_i) @>\partial >>
{\rm colim}\;{\rm Tor}_1(M,K_i) @>>>
0 \\
\end{CD}$$
and we get the desired isomorphism by conjugating the vertical isomorphism. One may show the isomorphism induced is natural, much like that of $M\otimes {\rm colim}$ and ${\rm colim}\; M\otimes $, but that is a bit more tortuous. Note this gives naturality of all the upper isomorphisms, since they are obtained by composing the natural connection morphisms and the natural isomorphism in the case $n=1$.
To use arbitrary quasi-isomorphisms $P^\bullet \to A^\bullet$, where $P^\bullet$ is a complex of projectives, you first need to show that such quasi-isomorphisms exist. That's what the Cartan-Eilenberg resolution accomplishes in the case when $A^\bullet$ is bounded above: it gives you a double complex of projectives such that its total complex is the complex $P^\bullet$ quasi-isomorphic to $A^\bullet$. Compare it to how we first show that any module over a commutative ring has a free resolution, even though we then use arbitrary projective resolutions in some computations.
The fact that a Cartan-Eilenberg resolution also induces resolutions of the homology of $A^\bullet$ is also sometimes useful, even though it's not needed to make sure hyperhomology is well-defined.
Best Answer
If $G$ is type $FP$ (or $FP_{\infty}$), then there is a resolution of $\mathbb{Z}$ by finitely generated projective $\mathbb{Z}G$-modules. (That they are finitely generated is the key property that you need.)
Suppose that $f \in Hom_G(P, \bigoplus M_i)$. Then there appears to be an induced element of $\bigoplus Hom(P, M_i)$ defined by $f_i(p) = (\pi_i \circ f)(p)$, where $\pi_i$ is the projection onto the $i$-th factor. The devil is in the details however. An element of $\bigoplus Hom(P, M_i)$ must have that all but finitely many $f_i$ are zero, whereas you are given that for each $p$, $f_i(p)$ is zero for all but finitely many $i$. But if $P$ is finitely generated, you can show that indeed all but finitely many $f_i$ are the zero map.