This is a standard cocycle averaging situation.
Choose an arbitrary (set-theoretic) section $s : G/N \to G$. The extent to which $s(G/N)$ fails to be a subgroup is measured by the cocycle $f : (G/N)^2 \to N$ defined by $$s(x)s(y) = s(xy) f(x,y) \qquad (x, y \in G/N).$$
This is indeed a cocycle in the sense that it satisfies the cocycle relation (using additive notation in $N \cong \mathbb{R}^n$) $$f(x,y)^z + f(xy,z) = f(x,yz) + f(y,z)$$
(this follows from expanding $s(x)s(y)s(z)$ in two different ways, using associativity). The exponent $z$ refers to the natural action of $G$ on $N$, which factors through $G/N$. Now we use the fact that $N$ is a vector space and $G/N$ is finite to define
$$F(y) = \frac{1}{|G/N|} \sum_{x \in G/N} f(x, y).$$
From the cocycle relation it follows that
$$F(y)^z + F(z) = F(yz) + f(y, z)$$
(i.e, $f$ is a coboundary). Hence, from the definition of $f$,
$$s(x)F(x)^{-1} s(y) F(y)^{-1} = s(xy) F(yz)^{-1},$$
so $x \mapsto s(x) F(x)^{-1}$ is a homomorphism $G/N \to G$, as required.
In homological terms this argument shows that $H^2(Q, M) = 0$ whenever $Q$ is a finite group and $M$ is a $Q$-module which is uniquely divisible by $|Q|$. More generally $H^n(Q, M) = 0$ for all $n > 0$ (Robinson, A course in the theory of groups, 11.3.8). For example also $H^1=0$, which means that, in your situation, all subgroups of $G$ of order $|G/N|$ are conjugate.
The more general case in which $G/N$ is compact is similar, using the Haar measure to average over $G/N$, assuming we have a continuous section $s : G/N \to G$. Maybe the existence of a continuous section is the difficult part.
