It seems like a homework question so I will not give full details.
First, the question asks to show that a straightforward application of DFT takes $\Theta(2^{2n})$ operations on an input with $2^n$ components. This is quite easy to see if we look at the $2^n \times 2^n = 2^{2n}$ matrix of DFT:
$$
W = \frac{1}{\sqrt{2^n}} \begin{bmatrix}
1&1&1&1&\cdots &1 \\
1&\omega&\omega^2&\omega^3&\cdots&\omega^{2^n-1} \\
1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(2^n-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(2^n-1)}\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&\omega^{2^n-1}&\omega^{2(2^n-1)}&\omega^{3(2^n-1)}&\cdots&\omega^{(2^n-1)(2^n-1)}
\end{bmatrix}.
$$
If we multiply $W$ with a vector and count the operations we get the result.
Now, how can we cut down the operation count using equation (5.4)?
Equation (5.4) allows you to take advantage of the fact that the Fourier transformed $|j_1, j_2, \ldots, j_n\rangle$ is made out of $n$ tensored $2\times 1$ vectors. So, we process each $2\times 1$ vector independently by performing the following $n$ mappings:
\begin{align}
\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) &\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + e^{2\pi i 0.j_n}|1\rangle),\\
\vdots\\
\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) &\rightarrow \frac{1}{\sqrt{2}}(|0\rangle + e^{2\pi i 0.j_1 \ldots j_n}|1\rangle).
\end{align}
Each mapping takes a constant number of operations in $n$ as it is simply multiplying a $2 \times 1$ vector by a $2\times 2$ phase matrix. I believe in the book the matrix is as follows:
$$
R_k = \begin{pmatrix}
1 && 0\\
0 && e^{2 \pi i /2^k}
\end{pmatrix}.
$$
Hence, we perform $n$ matrix-vector multiplication to process a single $|j_1\ldots j_n\rangle$.
We know that an arbitrary vector $|\psi\rangle$ on $n$ qubits can be written as a linear combination of $2^n$ binary kets $|j_1, j_2, \ldots, j_n\rangle$. For example, for $n=2$, an arbitrary state can be written as
a linear combination of $2^2$ binary kets like so
$$
|\psi\rangle = a |00\rangle + b|01\rangle +c|10\rangle+d|11\rangle.
$$
Therefore, to transform $|\psi\rangle$ on $n$ qubits, we need to process $2^n$ binary vectors $|j_1, \ldots j_n\rangle$ by performing $n$ mappings described above. Since each such binary vector requires $n$ matrix-vector multiplications, and there are $2^n$ of them, it takes $\Theta(n 2^n)$ operations.
