Here is a short proof that at most $2n-2$ swaps are necessary.  We proceed by induction on $n$.  For the base case $n=1$, it is clear that no swaps are necessary.  For the inductive step, let $X_1,X_2,X_3,X_4$ be multisets of $n$ numbers in $[0,1]$.  For each $i \in [4]$, arbitrarily choose $x_i \in X_i$, and let $Y_i=X_i \setminus \{x_i\}$.  By induction, we may perform at most $2n-4$ swaps to $Y_1 ,\dots, Y_4$ to obtain $Y_1', \dots, Y_4'$ such that $|\sigma(Y_i') - \sigma(Y_j')| \leq 1$ for all $i,j$.  By renaming, we may assume that $\sigma(Y_1') \leq \sigma(Y_2') \leq \sigma(Y_3') \leq \sigma(Y_4')$. For each $i \in  \{2,3,4\}$ let $d_i=\sigma(Y_i')-\sigma(Y_1')$.   It suffices to reorder $x_1, \dots, x_4$ using at most two swaps, such that $|\sigma(Y_i' \cup \{x_i)\}) - \sigma(Y_j' \cup \{x_j\})| \leq 1$ for all $i,j$.  By performing a single swap, we may assume that $x_1$ is the largest element among $x_1, \dots, x_4$. If $x_2 \geq \max\{x_3,x_4\}$, then by possibly swapping $x_3$ and $x_4$ (if necessary), we are done.  Thus, we may assume that $x_2 < \max\{x_3,x_4\}$.  Choose the index $i \in \{3,4\}$ such that $d_{i}+x_{i}$ is maximum.  Swapping $x_i$ with $x_2$ gives the required reordering of $x_1, \dots, x_4$.  $\Box$

By choosing $x_1, \dots, x_4$ more cleverly, or strengthening the induction hypothesis, it might be possible to obtain a better bound.