What's the usual $G$- action on group extension? From Relation Modules Of Finite Groups by Gruenberg:

Let $1\rightarrow K\rightarrow E\xrightarrow{\text{$\pi$}} G\rightarrow 1$ be exact sequence of groups, $G$ finite and $K\lhd E $.
Denote $\Delta K$ the augmentation ideal of $K$ (kernel of the map $\mathbb Z K\rightarrow \mathbb Z$ by $\sum m_k k\mapsto \sum m_k$.), then for any (right) $E$-module $M$, $M\Delta K$ is a submodule and $M/(M\Delta K)$ is obviously a $G$-module.
($M/(M\Delta K)$ can also be expressed as $M\otimes_{\mathbb ZE} \mathbb Z G$.)

How is $G$-action defined on $M/(M\Delta K)$? And how $M/(M\Delta K)$ can be expressed as $M\otimes_{\mathbb ZE} \mathbb Z G$ as the author said?
 A: Here's my answer to the first question (I wouldn't call it obvious).
For $g\in G$ define 
$$(m+M\Delta K)g:=me+M\Delta K$$ for some $e\in\pi^{-1}(\{g\})$. This does not depend on the choice of $e$.
Indeed, if $\pi(e)=\pi(f)$ then $ef^{-1}\in K$ by exactness. Now apply the augmentation map $A:\mathbb{Z}K\to\mathbb{Z}$ to get $A(ef^{-1})=1$ and so $ef^{-1}-1\in \Delta K$. This means $m(ef^{-1}-1)\in M\Delta K$ and since $M\Delta K$ is a $E$-submodule (due to $K$ being normal in $E$) then you mulitply by $f$ to get $m(e-f)\in M\Delta K$ meaning $me+M\Delta K=mf+M\Delta K$. Proving that the action is well defined.

So what about the other question? Let's start with the standard (right) $\mathbb{Z}E$ action on $\mathbb{Z}G$ given by $g\cdot e:=g\pi(e)$. This has some interesting properties on the tensor product, namely 
$$me\otimes 1=m\otimes \pi(e)$$
And since $\pi$ is onto then every element of the tensor product is of the form $m\otimes 1$. Note that $me\otimes 1=mf\otimes 1$ if $\pi(e)=\pi(f)$. Sounds familiar? Even better, this gives us a natural way to define a $G$-module structure on $M\otimes\mathbb{Z}G$, namely
$$(m\otimes 1)\cdot g:=m\otimes g=me\otimes 1$$
for $e\in\pi^{-1}(\{g\})$. 
Finally we have a morphism $M\to M\otimes\mathbb{Z}G$ given by $m\mapsto m\otimes 1$. And what we've discussed so far shows that this is an epimorphism. Next step is to use the first isomorphism theorem for $\mathbb{Z}E$ modules and to show that the induced isomorphism preserves $\mathbb{Z}G$ action (remember that $M$ is not a $\mathbb{Z}G$ module) which in this case should be trivial. Calculating the kernel is a bit challenging though. But should be similar to the first section.
