(I write an answer rather than a comment in order to accommodate exact sequences.) 

Let 
$$0\to T\to E\to\Gamma\to 1\tag{$E_1$}$$ 
be your first  group extension, where $T$ is the torus ${\Bbb R}^n/{\Bbb Z}^n$. 
Write $R_k\subset T$ for the kernel of multiplication by $k$ in $T$ 
and consider your second exact sequence
$$0\to T/R_k\to E/R_k\to\Gamma\to 1.\tag{$E_2$}$$ 
To the  extension $(E_1)$ we associate its cohomology class $\eta_1\in H^2(\Gamma,T)$, and to extension $(E_2)$ we associate its class $\eta_2\in H^2(\Gamma,T/R_k)$.
Then it follows from the constructions of $\eta_1$ and $\eta_2$ that $\eta_2$ is the image of $\eta_1$ under the homomorphism
$$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T/R_k)$$
induced by the canonical homomorphism 
$$\phi\colon T\to T/R_k.$$
We have not yet used the assumption that $T$ is a torus and that $\#\Gamma=k$.

Now consider the *surjective* homomorphism
$$\alpha\colon  T\to T,\quad x\mapsto kx.$$
Its kernel is $R_k$, and so it induces an isomorphism 
$$\alpha_*\colon T/R_k\to T.$$
Identifying $T/R_k$ with $T$ using $\alpha_*$, we obtain that our 
$$\phi\colon T\to T$$ 
is multiplication by $k$.
It follows that 
$$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T)$$
is multiplication by $k$ as well.
Since $\Gamma$ is a group of order $k$, multiplication by $k$ annihilates $H^2(\Gamma,T)$.
See Corollary 1 of Proposition 8 in Section 6, page 105, of: 
Atiyah and Wall, Cohomology of groups, in: Cassels and Fröhlich (eds.), Algebraic Number Theory, Acad. Press 1967, pp. 94-115. It follows that $\eta_2=0$ and the sequence $(E_2)$ splits.