Complexity of FFT derivation
No, Equation (2) is not the sum of 2 FFTs (of lengths $N/2$) as you claim, and it is not computationally more efficient. The canonical way of evaluating Equation (1) is via Horner's method:

$$X[k] = (((\cdots (x[N-1]\cdot\omega + x[N-2])\cdot\omega + x[N-3])\cdot\omega + \ldots)\cdot\omega + x[0]\tag{a}$$

where $\omega = W_N^k$ which requires $N-1$ multiplications and $N$ additions: lather, rinse, and repeat for for other values of $k$. Equation (2) just splits this Horner's rule computation into two parts with no change in complexity. That is, we compute $\sum_{n\ even}x[n]W_{N}^{nk}$  as $$(((\cdots (x[N-1]\cdot\Omega + x[N-3])\cdot\Omega + x[N-5])\cdot\Omega + \ldots)\cdot\Omega + x[0]\tag{b}$$ where $\Omega = \omega^2$ and
$\sum_{n\ odd}x[n]W_{N}^{nk}$ as 
$$(((\cdots (x[N-2]\cdot\Omega + x[N-4])\cdot\Omega + x[N-6])\cdot\Omega + \ldots)\cdot\Omega + x[1])\cdot\omega.\tag{c}$$ Each computation requires $N/2$ multiplications and $N/2$ additions with no savings in computation over the standard brute-force Horner's method applied to a vector of length $N$ in $(a)$.

Equation (3), however, is the core of the FFT butterfly; the two sums in Equation (3) are DFTs of two shorter length sequences.  Define \begin{align}x_{\text{even}}[k] &amp;= x[2k], &amp;k = 0, 1, \ldots, N/2-1,\\
x_{\text{odd}}[k] &amp;= x[2k+1], &amp;k = 0, 1, \ldots, N/2-1,
\end{align} and let the DFTs of these shorter sequences be denoted by $X_{\text{even}}[k]$ and $X_{\text{odd}}[k]$ respectively. Then the computation of Equation (1) has been re-written in Equation (3) as
$$X[k] = X_{\text{even}}[k] + W_N^kX_{\text{odd}}[k] \tag{**}$$
which requires one multiplication and one addition per bin. Note that $k$ varies from $0$ to $N-1$ in $(**)$, and the shorter DFTs in $(**)$ are assumed to be periodic functions of period $N/2$ (where's rb-j when I need him?)  But, now note that $X_{\text{even}}[k]$ and $X_{\text{odd}}[k]$ are DFTs of length $N/2$, and so, if $N/2$ is an even number (hint: this means that $N$ is a multiple of $4$), we can use the idea behind $(**)$ to express $X_{\text{even}}[k]$ as
$$X_{\text{even}}[k] =  X_{\text{even, even}}[k] + W_{N/2}^kX_{\text{even, odd}}[k]$$
and $X_{\text{odd}}[k]$ as
$$X_{\text{odd}}[k] =  X_{\text{odd, even}}[k] + W_{N/2}^kX_{\text{odd, odd}}[k]$$
that is, both DFTs of length $N/2$ have been expressed in terms of 2 DFTs of length $N/4$, and so on.  If $N = 2^m$, we can proceed in this way to where we have $N/2$ DFTs of length $2$ that we compute using one multiplication and addition per DFT bin. So, $N$ multiplications and additions total to compute these $N/2$ DFTs. (Purists whining that a length $2$ DFT needs only one addition and one subtraction are kindly requested to go jump in the lake!). Next, we combine these $N/2$ DFTs of length $2$ into $N/4$ DFTs of length $4$ using one multiplication and one addition per DFT bin for a total of $N$ multiplications and additions to do the combination. Next, we combine these $N/4$ DFTs of length $4$ into $N/8$ DFTs of length $8$ using one multiplication and one addition per DFT bin for a total of $N$ multiplications and additions to do the combination..... Ultimately, we arrive at $(**)$.  This iterative procedure for computing the DFT of length $N$ is (one of the many versions of) the FFT algorithm for computing the Discrete Fourier transform. It has $m=\log_2N$ stages requiring $N$ multiplications and $N$ additions at each stage for a total of $Nm = N\log_2N$ multiplications and additions. In contrast, the use of Horner's rule would need $N(N-1)$ multiplications and $N^2$ additions. Hence the soubriquet FAST for this kind of algorithm.

Purists who have survived their jumps into the lake can now start their cacophony about radix-2 versus radix-4 versus radix-8 transforms and the exact number of multiplications and additions needed for implementing an FFT algorithm.