As far as I understand, 2X-Thru deembedding works by directly connecting the two halves of the test fixture together, doing a two-port measurement of the result, and using this result to deduce the parameters of each half of the test fixture.
We start from the generic equations for the concatenation of two networks described by S parameters,
$$
\begin{eqnarray*}
S_{11} &=& S_{11A} + \frac{S_{11B} S_{21A} S_{12A}}{1 - S_{22A} S_{11B}}\\
S_{12} &=& S_{21A} \frac{S_{21B}}{1 - S_{22A} S_{11B}}\\
S_{21} &=& S_{12A} \frac{S_{12B}}{1 - S_{22A} S_{11B}}\\
S_{22} &=& S_{22B} + \frac{S_{22A} S_{21B} S_{12B}}{1 - S_{22A} S_{11B}}
\end{eqnarray*}
$$
assume that each half of the test fixture is described by S parameters $S_{11f}$, $S_{12f}$, $S_{21f}$, $S_{22f}$; and that the right half is an exact mirror image of the left half,
$$
\begin{eqnarray*}
S_{11A} &=& S_{11f}\\
S_{12A} &=& S_{12f}\\
S_{21A} &=& S_{21f}\\
S_{22A} &=& S_{22f}
\end{eqnarray*}
$$
$$
\begin{eqnarray*}
S_{11B} &=& S_{22f}\\
S_{12B} &=& S_{21f}\\
S_{21B} &=& S_{12f}\\
S_{22B} &=& S_{11f}
\end{eqnarray*}
$$
to arrive at
$$
\begin{eqnarray*}
S_{11} &=& S_{11f} + \frac{S_{22f} S_{21f} S_{12f}}{1 - S_{22f} S_{22f}}\\
S_{12} &=& \frac{S_{21f} S_{12f}}{1 - S_{22f} S_{22f}}\\
S_{21} &=& \frac{S_{12f} S_{21f}}{1 - S_{22f} S_{22f}}\\
S_{22} &=& S_{11f} + \frac{S_{22f} S_{12f} S_{21f}}{1 - S_{22f} S_{22f}}\\
\end{eqnarray*}
$$
From this, we see that $S_{11} = S_{22}$ and $S_{12} = S_{21}$, i.e. the S parameters for the complete fixture are symmetrical. (This is also intuitively clear: if the right half of the fixture is a mirror image of the left half, mirroring the entire fixture should not change anything.) Thus, if we make no further assumptions about the parameters $S_{ijf}$, we effectively end up with two equations for four unknowns.
The IEEE 370 standard appears to recognize this problem and states (in D.6.1): "The assumptions are the following: [...] Each half of the 2X-Thru is symmetric." Under this assumption, we have:
$$
\begin{eqnarray*}
S_{22f} &=& S_{11f}\\
S_{21f} &=& S_{12f}\\
\end{eqnarray*}
$$
and thus two equations for two unknowns:
$$
\begin{eqnarray*}
S_{11} &=& S_{11f} + \frac{S_{11f} S_{12f}^2}{1 - S_{11f}^2}\\
S_{12} &=& \frac{S_{12f}^2}{1 - S_{11f}^2}
\end{eqnarray*}
$$
Under the assumption that $(1 - S_{11f}^2) \neq 0$, these can be solved to give:
$$
\begin{eqnarray*}
S_{11f} &=& \frac{S_{11}}{1 + S_{12}}\\
S_{12f}^2 &=& S_{12} (1 - S_{11f}^2)
\end{eqnarray*}
$$
How valid the symmetry assumption for each half of the fixture is is IMHO debatable.
