Defining the exchange operator $P_{12}$ without reference to dummy variables Let $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ so that $\Psi$ represents the state of two interacting fermions.
Traditionally, we define the exchange operator $P_{12}$ via
\begin{align*}
P_{12}\Psi(\mathbf{r}_1, \mathbf{r}_2) = \Psi(\mathbf{r}_2, \mathbf{r}_1)
\end{align*}
but pretty soon we realize this is incorrect, as we know that we need to interchange the spin state as well. This is often achieved by writing
\begin{align*}
P_{12}\Psi(\mathbf{r}_1, \sigma_1, \mathbf{r}_2, \sigma_2) = \Psi(\mathbf{r}_2, \sigma_2, \mathbf{r}_1, \sigma_1)
\end{align*}
but this is still awkward, as the spin state is a dependent variable in the theory.
To avoid these problems I would like to define the exchange operator without any reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$.
This is what I have so far: Suppose $\{\psi_{n}\}$ separates $\mathcal{L}^{2}(\mathbb{R}^{3}, \mathbb{C})$.
Then any function $\Psi \in \mathcal{L}^{2}(\mathbb{R}^{6}; \mathbb{C}^{2}\otimes\mathbb{C}^{2})$ can be written as
\begin{align}\label{separate}
\Psi
&=
\sum_{n} a_{nm}
\begin{pmatrix}
\psi_{n} \\
0
\end{pmatrix}
\otimes
\begin{pmatrix}
0 \\
\psi_{m}
\end{pmatrix}
+
b_{nm}
\begin{pmatrix}
\psi_{n} \\
0
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{m} \\
0
\end{pmatrix}
+
c_{nm}
\begin{pmatrix}
0 \\
\psi_{n}
\end{pmatrix}
\otimes
\begin{pmatrix}
0 \\
\psi_{m}
\end{pmatrix}
+
d_{nm}
\begin{pmatrix}
0 \\
\psi_{n}
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{m} \\
0
\end{pmatrix}
\end{align}
where the first part of the tensor product is assumed to act on $\mathbf{r}_1$ and the second assumed to act of $\mathbf{r}_2$.
Now we define $P_{12}$ on the basis, interchanging the functions rather than the dummy variables.
Exchange of the spin-coordinate is now trivial, defined by
\begin{align*}
P_{12}^{\mathrm{spin}}
\begin{pmatrix}
\psi_{n} \\
\psi_{m}
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{j} \\
\psi_{k}
\end{pmatrix}
=
\begin{pmatrix}
\psi_{m} \\
\psi_{n}
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{k} \\
\psi_{j}
\end{pmatrix}
\end{align*}
However, what I would consider the natural space exchange definition
\begin{align*}
P_{12}^{\mathrm{space}}
\begin{pmatrix}
\psi_{n} \\
\psi_{m}
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{j} \\
\psi_{k}
\end{pmatrix}
=
\begin{pmatrix}
\psi_{j} \\
\psi_{k}
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi_{n} \\
\psi_{m}
\end{pmatrix}
\end{align*}
gives results which are in disagreement with the notation found in standard quantum mechanics texts. For instance, take $\Psi(\mathbf{r}_1, \mathbf{r}_2) = \psi(\mathbf{r}_1)\psi(\mathbf{r}_2)(\uparrow \downarrow - \downarrow \uparrow)$. This is clearly antisymmetric, but is symmetric under $P_{12}^{\mathrm{space}}P_{12}^{\mathrm{spin}}$. This can be patched up by defining $P_{12}^{\mathrm{space}}$ differently on each of the four types of terms is \ref{separate}, but that seems so unnatural that I feel like I'm on the wrong track.
Is there a natural definition of $P_{12}$ which doesn't make reference to the dummy variables $\mathbf{r}_1$ and $\mathbf{r}_2$?

The column vector notation can be used to express the result of exchange of spatial or spin arguments of the psi function, if we adopt this convention:
if the symbol is on the left side of the tensor product, it belongs to particle 1, if it is on the right side, it belongs to particle 2.
This means that when the symbol $\psi_n$ is moved from the left vector to the right vector as part of your operation $P_{12}^{space}$, the particle index of the spatial argument changes from 1 to 2 which is desired, but the symbol then multiplies spin vector of 2nd particle, not spin vector of the 1st one as it should. Similar problem occurs for your operator $P_{12}^{spin}$.
Consequently, one cannot represent the exchange operations for spatial or spin arguments in the way you wrote.
Using the original function notation it can be derived that the exchange operators work this way in the column vector notation:
\begin{align}
P_{12}^{space} \begin{pmatrix}
f_+ \\
f_-
\end{pmatrix}
\otimes
\begin{pmatrix}
g_+ \\
g_-
\end{pmatrix} =
\end{align}
\begin{align}
=
\begin{pmatrix}
g_+ \\
.
\end{pmatrix}
\otimes
\begin{pmatrix}
f_+ \\
.
\end{pmatrix}
+
\begin{pmatrix}
g_- \\
.
\end{pmatrix}
\otimes
\begin{pmatrix}
. \\
f_+
\end{pmatrix}
+
\begin{pmatrix}
. \\
g_+
\end{pmatrix}
\otimes
\begin{pmatrix}
f_- \\
.
\end{pmatrix}
+
\begin{pmatrix}
. \\
g_-
\end{pmatrix}
\otimes
\begin{pmatrix}
. \\
f_-
\end{pmatrix}.
\end{align}
These 4 terms cannot be simplified into lesser number of terms, because they are linearly independent. For spin exchange the result is similar, just exchange $f$ with $g$ everywhere.
You can then try to verify that subsequent application of $P_{12}^{space}$ and $P_{12}^{spin}$ on the ket
\begin{align}
\begin{pmatrix}
\psi\\
.
\end{pmatrix}
\otimes
\begin{pmatrix}
. \\
\psi
\end{pmatrix}
-
\begin{pmatrix}
. \\
\psi
\end{pmatrix}
\otimes
\begin{pmatrix}
\psi\\
.
\end{pmatrix}
\end{align}
results in minus this ket.
