The discrepancy comes from the fact that there are different ways to define a quantum Fourier transform and the direct equivalence to the Hadamard transform holds for only one such definition. See for instance this Introduction to Quantum Computing.
Consider the general case of $N$ qubits, let the $2^N$ computational basis states be
$$
|{\bf x}\rangle \equiv |x_1, x_2, x_3, ... ,x_N\rangle = |x_1\rangle \otimes |x_2\rangle \otimes |x_3\rangle \otimes ...|x_N\rangle \;\;\;\text{for} \;\;\; x_i = 0 \; \text{or} \; 1
$$
and write an arbitrary state as
$$
|f \rangle = \sum_{{\bf x}}{f({\bf x})|{\bf x}\rangle}
$$
The general definition of a quantum Fourier transform is
$$
|\hat{f}\rangle = \sum_{{\bf x}}{\hat{f}({\bf x})|{\bf x}\rangle}
$$
where
$$
\hat{f}({\bf x}) = \frac{1}{2^{N/2}}\sum_{{\bf y}}{\chi_{{\bf y}}({\bf x}) f({\bf y})}
$$
and for every ${\bf x}$ the numbers $\chi_{{\bf y}}({\bf x})$ are distinct $2^N$-th complex roots of the unit, $\left[\chi_{{\bf y}}({\bf x})\right]^{\left(2^N\right)} = 1$. The particular correspondence ${\bf x}, {\bf y} \rightarrow \chi_{{\bf y}}({\bf x})$ is formally introduced by means of a group representation on the computational basis states and the $\chi$-s represent group characters.
The choice of $\chi_{{\bf y}}({\bf x})$ for which the quantum Fourier transform is identical to a Hadamard transform corresponds to a direct discrete Fourier transform on the amplitudes $f({\bf x})$,
$$
\chi_{{\bf y}}({\bf x}) = (-1)^{{\bf x} \cdot {\bf y}}\\
\hat{f}({\bf x}) = \frac{1}{2^{N/2}}\sum_{{\bf y}}{(-1)^{{\bf x} \cdot {\bf y}} f({\bf y})}
$$
In this case the transformation is a direct product of $N$ Hadamard transforms, $H_{2^N} = H_2\otimes H_2 \otimes ... \otimes H_2$. For $N=2$ this is $H_4 = H_2\otimes H_2 $.
But the choice that reproduces the quantum Fourier transform in the question is
$$
\chi_{{\bf y}}({\bf x}) = e^{i\frac{2\pi\beta({\bf x})\beta({\bf y})}{2^N}} \\
\hat{f}({\bf x}) = \frac{1}{2^{N/2}}\sum_{{\bf y}}{e^{i\frac{2\pi\beta({\bf x})\beta({\bf y})}{2^N}} f({\bf y})}
$$
which uses the binary map ${\bf x} \rightarrow \beta({\bf x}) \in \mathbb{N}$,
$$
\beta({\bf x}) = \sum_{k=0}^{N-1}{2^kx_k}
$$
In this case the Hadamard transforms on individual qubits are each followed by a cascade of controlled phase-shifts on subsequent qubits:
For 2 qubits this becomes the circuit and matrix here.
