I got stuck on the following:
Let $G$ be a finite cyclic group. Then it is a well-known fact, that one can compute its Tate-cohomology groups from the complex $$\cdots\xrightarrow{\;\tau \;-\;\text{Id} \;}M \xrightarrow{\;\;Tr_{G}\;\;} M \xrightarrow{\;\tau \;-\;\text{Id} \;} M \xrightarrow{\;\;Tr_G\;\;} M \xrightarrow{\;\tau \;-\;\text{Id} \;}\cdots $$ where $\tau$ is multiplication by some generator of $G$ and $Tr_G(x)= \sum_{g \in G} gx$.
Can you tell me why this defines an acyclic resolution of our module? I do not see why this complex necessarily has to compute the cohomology groups.
It doesn't define an acyclic resolution (at least, unless $M$ is acyclic). But one can show that the cohomology groups that arise from that complex are isomorphic to the usual ones as follows:
The following is from Chapter VIII, $\S$4 of Serre's Local Fields (I changed the notation a bit to fit yours though):
The last bit is using that a $G$-module is relatively projective if and only if it is relatively injective when $G$ is finite, and an abstract characterization of derived functors (of which group cohomology is an example) - see the last sentence of the section here.