I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis?
Having experienced on a number of occasions how counter-intuitive rotational dynamics can be I will not be surprised to find that my answer is flawed.
The period of revolution of the Moon about its own axis is the same as its period of revolution about the Earth.
This is due to tidal locking.
Diagram $(1)$ (adapted from the diagram in Wikipedia article "Tidal force") has the Earth somewhere below the diagram and shows the Earth's gravity differential field at the surface of the Moon which causes the Moon to change its shape.
Under the influence of these differential forces the Moon changes its shape to something like the greatly exaggerated green ellipsoid shown in diagram $(2)$.
There is no net torque on the Moon due to the Earth and so its period of revolution about its axis stays the same.
The Wikipedia article Tidal locking has a diagram $(2)$ which shows what the differential forces on the Moon would have been like if the Moon's period of revolution about its axis was greater than its period of revolution about the Earth.
There is a net torque on the Moon which reduces its spin angular momentum.
Ignoring the influence of the Sun and the rest of the Solar System as there are no external torques on the Moon-Earth system the loss in spin angular momentum of the Moon must increase the angular momentum of the system somewhere else. It is the orbital angular momentum of the Moon which increases as does, to a lesser extent, the spin angular momentum of the Earth.
Both Wikipedia articles are very informative.
The question is, "Where is the force that increases the orbital angular momentum?"
I think that the answer to the original question is, the force $F$ in diagram $(5)$?
The diagrams I have drawn are gross exaggerations of what actually happens. Any smaller and the force $F$ would be difficult to identify. I have also only considered a situation where all the mass is distributed in a plane and ignored any angular momentum components which are not perpendicular to that plane.
In diagram $(3)$ the Earth $E$ is at the bottom and the Moon's centre of mass $M$ is orbiting the Earth with an angular speed $\omega$.
The Moon is rotating about its centre of mass with an angular speed $\Omega (>\omega)$.
$F_c$ is the gravitational attraction due to the Earth on a particle at the centre of mass of the Moon.
$F_f$ and $F_n$ are the gravitational attractive forces acting on particles at points $A$ and $B$ on the Moon.
The differential forces acting on the particles at $A$ and $B$ are $F_f^\prime$ and $F_f^\prime$.
The diagram shows that these forces, which are responsible for reducing the spin angular momentum of the Moon, are not parallel to one another nor do they have the same magnitude.
Hence there is also a net force $F$ which acts at the centre of mass of the Moon.
Diagram $(4)$ illustrate the fact that for a body which is symmetrical about the $EM$ axis there is no torque.
If there is asymmetry about the axis $EM$ as in diagram $(2)$ then tidal locking is possible.
Diagram $5$ shows the couples (in green and blue) acting on the Moon which reduce the Moon's spin angular momentum and the net force $F$ (in red) which increases the Moon's orbital angular momentum about the Earth.