A general decomposition of a $\bigwedge_2U$ gate (where U is unitary) into single-qubit gates and $C_{not}$ gates can be found as Lemma 6.1 in Barenco et al. "Elementary gates for quantum computation", Physical Review A, 52(5):3457–3467, 1995:
where $V$ is chosen such that $V^2 = U$. Such a $V$ exists, since any linear transformation can be brought into upper triangular form by the Schur decomposition, and if $U$ is unitary, its rows and columns are orthonormal, hence it must be diagonal. Then simply take $V$ to be diagonal with square roots of the diagonal entries of $U$. Furthermore, it is seen that $V$ is also unitary.
Now, the above circuit maps the control bits and a generic bottom bit $\ket x$ as follows
(1)
\begin{align} \ket{00x} &\mapsto \ket{00x} \\ \ket{10x} &\mapsto \ket{10V^{\dagger}V(x)} = \ket{10x} \\ \ket{01x} &\mapsto \ket{01VV^{\dagger}(x)} = \ket{01x} \\ \ket{11x} &\mapsto \ket{11VV(x)} = \ket{11U(x)} \end{align}
That is, it implements $\bigwedge_2U$.
If
(2)
\begin{align} U = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{align}
then $X$ can be diagonalised to
(3)
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
in the basis $\tfrac1{\sqrt 2} (\ket 0 + \ket 1),\, \tfrac1{\sqrt 2} (\ket 0 - \ket 1)$.
Hence, we can take $V$ in $X = V^2$ as
(4)
\begin{align} V = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} \end{align}
and apply the construction in section 5.4.2 for $\bigwedge V$. This supplies the simple transformations required for $\bigwedge_2X$.
Note, that by the change of basis transformation
(5)
\begin{align} B = \frac1{\sqrt 2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{align}
the expression for $V$ in the standard basis $\ket 0,\, \ket 1$ is
(6)
\begin{align} B^{-1} \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} B = \frac{1}{2} \begin{pmatrix} 1+i & 1-i \\ 1-i & 1+i \end{pmatrix}\,. \end{align}
