Theorem [LM2007a], [LM2007b]: If G has some smallest vertex-edge covering S of size τ(G) containing two vertices with at most one common neighbor, then the transfer can be done with a small boat of size τ(G) in at most 2·(n-τ(G))+3 moves.
Let x and y be these two vertices in S with at most one common neighbor. In the first move, S is loaded on the boat, and most of S stays on the boat most of the time. Actually after the first move, x is unloaded on the right bank, and the boat returns with the rest of S. Then the neighbors of y that are not neighbors of x are transferred one by one. If x and y have a common neighbor, u, this vertex is eventually moved as well, but when u is loaded, on the left bank, y is unloaded on the right bank. On the right bank, u is unloaded and x is loaded. Returning to the right bank, all other vertices are transferred one by one, with y as the last vertex.
Each time the boat leaves the right bank, the number of vertices there has just been increased by one, except when x is exchanged for u. At the last arrival of the boat at the right bank, the number of vertices there increases by τ(G). Therefore the number of moves needed is 2·(n-τ(G)+1)+1 = 2·(n-τ(G))+3 if x and y have one common neighbor, and 2·(n-τ(G))+1 if x and y have no common neighbor.
Corollary: [LM2007a], [LM2007b]: All graphs G with τ(G) > 1 without 4-cycles, and in particular all trees are small boat graphs, allowing transfer with at most 2·(n-τ(G))+3 moves.
Try this method in the tree below to transfer G in 21 moves. However, the method described above is nnot always optimal, since in the example below 17 moves suffice.