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We now see that $F_e(k)$ consists of even terms of $F(k)$, and $U_N^kF_o(k)$ consists of odd terms of $F(k)$. QED.
A Detail of Implementation
Originally $F_e(k)$ and $ F(k)$are both arrays of length $N/2$. However, in the last step we need to add the corresponding element from both into $F(k)$ of length $N$.
The key is the fact that $F_e(k)$ and $F_o(k)$ are both $N/2$ periodical. So, $F_e(k) = F_e(k - N/2)$. This also holds for $F_o(k)$.
We can calculate $F(k)$ like this:
int M = N / 2;
for (int k = 0; k < N; k++) {
F[k] = Fe[k % M] + RU(k, N) * Fo[k % M];
}