$(2n-1)!!$ is the number of ways to divide $2n$ distinct items into pairs, so $(2n-3)!!$ is the number of ways to pair off $2n-2$ items. Thus, it’s natural to try to find a bijection between binary total partitions of $[n]$ and partitions of $[2n-2]$ into pairs.
There is an obvious bijection between binary total partitions of $[n]$ and full binary trees whose leaves are labelled by the singletons of the elements of $[n]$. Such a tree has $n-1$ internal vertices, including the root, each of which has exactly two children. Thus, there are $n-1$ pairs of children. Somehow we need to label these children with elements of $[2n-2]$ in such a way that the tree (and hence the binary total partition) can be reconstructed from the pairs.
The root is not a leaf, so every leaf is a child. Each leaf already bears a label of the form $\{k\}$ for some $k\in[n]$, and we can simply replace that label with the label $k$. That accounts for the labels in $[n]$, leaving only those in $[2n-2]\setminus[n]=\{n+1,\ldots,2n-2\}$ to be placed in the tree. The problem, of course, is that we have to do this without knowing exactly what the tree looks like. This suggests that we should work back up the tree from the leaves, assigning a label to each interior vertex only after we’ve assigned labels to both of its children. We’ll do this recursively; the labelling of the leaves gets the recursion started.
Call an interior vertex eligible if both of its children have been assigned labels from $[2n-2]$. (I’ll refer to these as the new labels to distinguish them from the old labels, i.e., those that are subsets of $[n]$ appearing in the binary total partition of $[n]$.) At each stage of the recursion consider the set of eligible vertices. Each of those has an old label that is a subset of $[n]$, and these subsets are pairwise disjoint, since an eligible vertex cannot be a descendant of another eligible vertex. Pick the eligible vertex whose old label has the smallest minimum element, and assign to it the smallest new label that has not yet been assigned. Continue until all non-root vertices have received new labels. (Note that the root will be the last vertex to become eligible and will do so only when all other vertices have been relabelled, so we need never assign it a new label.) We can now assign to the tree (and hence to the original binary total partition of $[n]$ the set of (unordered) pairs of new labels of sibling vertices.
Consider, for example, the binary total partition of $[7]$ organized in tree form below:
{1,2,3,4,5,6,7}
{2,5} {1,3,4,6,7}
{2} {5} {3,6,7} {1,4}
{6} {3,7} {1} {4}
{3} {7}
The corresponding tree, after the leaves have been relabelled:
*
/ \
/ \
/ \
* *
/ \ / \
/ \ / \
2 5 * *
/ \ / \
6 * 1 4
/ \
3 7
The eligible vertices are those with the old labels $\{2,5\},\{3,7\}$, and $\{1,4\}$. Of these, the one with the smallest minimum member is $\{1,4\}$, so it gets the new label $8$. The only eligible vertices now are those with the old labels $\{2,5\}$ and $\{3,7\}$; the former has the smaller minimum element and so gets the new label $9$. This still creates no new eligible vertex: the only one left is $\{3,7\}$, which gets the new label $10$.
This makes the vertex $\{3,6,7\}$ eligible; since it’s the only eligible vertex, it gets the new label $11$. Vertex $\{1,3,4,6,7\}$ is now the sole eligible vertex and gets the new label $12$; this completes the relabelling, and we assign to the binary total partition the set $$\big\{9,12\},\{2,5\},\{8,11\},\{6,10\},\{1,4\},\{3,7\}\big\}$$ of pairs of elements of $[12]$.
*
/ \
/ \
/ \
9 12
/ \ / \
/ \ / \
2 5 11 8
/ \ / \
6 10 1 4
/ \
3 7
It remains to be verified that the tree (and hence the partition) can be reconstructed from the set of pairs. Let $\mathscr{P}$ be any partition of $[2n-2]$ into pairs. Even if every member of $[2n-2]\setminus[n]$ is paired with a member of $[n]$, there must be at least one pair in $\mathscr{P}$ that is a subset of $[n]$. Each such pair corresponds to a pair of sibling leaves in the tree, and we start the reconstruction of the tree with these pairs. The parents of these sibling pairs were the eligible vertices at the first step of the relabelling (after the leaves were relabelled). The old labels of these parent vertices were simply two-element subsets of $[n]$ whose elements were the (new) labels of their leaf children. The one with the smallest minimum element got label $n+1$ in this step of the recursive relabelling, and we also know its old label.
Now look at all pairs whose elements are in $[n+1]$ and whose parents have not yet been added to the tree. This includes any other pairs of leaves and the pair $\{k,n+1\}$, if $k\in[n]$. The vertices corresponding to these pairs were the eligible vertices at the next stage of the original recursion, and their old labels are simply the unions of the old labels of their children, which are known. Thus, we know which one would have been chosen (by virtue of having the smallest minimum element of its old label); that vertex must have received the new label $n+2$. Clearly we can continue this procedure to reconstruct the tree and binary total partition.
For example, if we start with $n=7$ and the partition $$\big\{\{1,10\},\{2,9\},\{3,8\},\{4,7\},\{5,6\},\{11,12\}\big\}$$ of $[12]$ into pairs, we begin the reconstruction with the leaf pairs $\{4,7\}$ and $\{5,6\}$, whose parents have the old labels $\{4,7\}$ and $\{5,6\}$, respectively. These were the eligible vertices initially, and it’s $\{4,7\}$ that must have got the new label $8$. $3\in[7]$, so the pair $\{3,8\}$ is now eligible along with $\{5,6\}$ and in fact is the next one chosen: it must have $9$ as its new label. Note that since the old labels corresponding to new labels $3$ and $8$ are $\{3\}$ and $\{4,7\}$ respectively, the old label of vertex $9$ is $\{3,4,7\}$. The eligible pairs are now $\{2,9\}$ and $\{5,6\}$, and $2<5$, so $\{2,9\}$ must have the new label $10$ and the old label $\{2,3,4,7\}$.
Continuing in this fashion, we see that the pair $\{1,10\}$, corresponding to the old label $\{1,2,3,4,7\}$, has new label $11$, and the leaf pair $\{5,6\}$ finally gets new label $12$ (and of course old label $\{5,6\}$. Thus, the original partition was into the sets $\{1,2,3,4,7\}$ and $\{5,6\}$. The former was split into $\{1\}$ and $\{2,3,4,7\}$, and the latter, of course, into $\{5\}$ and $\{6\}$. $\{2,3,4,7\}$ was split into $\{2\}$ and $\{3,4,7\}$, and the latter finally into $\{3\}$ and $\{4,7\}$.
This correspondence provides the desired bijection. It’s a little awkward to describe, but it’s conceptually fairly simple.
