Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.
So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is nonzero
Let $G = C_4 \times C_2$ be the direct product of cyclic groups of orders $4$ and $2$, with the direct factors generated by elements $g$ and $h$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.
I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.
Consider the group defined by the following presentation. (I am putting it in Magma format for ease of cutting and pasting.)
E := Group< g, h, a, b | g^4=a*b, h^2=a^2=b^2=1, a^g=b, b^g=a,
a^h=a, b^h=b, a*b=b*a, h^g=h*a >;
You can check by computer that $|E|=32$ (it is $\mathtt{SmallGroup}(32,7))$, and you can see directly from the presentation that it is an extension of $M$ by $G$ with the prescribed induced module action of $G$ on $M$.
Now the extension corresponding to the image of the corresponding element of $H^2(G,M)$ under the induced map $H^2(G,M) \to H^2(G,T)$ is
Group< g, h, t | g^4=1, h^2=t^2=1, t^g=t, t^h=t, h^g=h*t >;
which defines a nonabelian group of order $16$, so it cannot be the split extension of $T$ by $G$.
There is a similar example with $G$ dihedral of order $8$.