Velocity in power calculations in different inertial frames In calculating power using the formula $\underline{F}\cdot\underline{v}$, what is the correct velocity to use? Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force? I always thought it was the former (at least because in the case of force fields the field doesn't have a velocity, so the only velocity is that of the body the force is acting on).
However, when I use this understanding on an example problem I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of. I have posted this question here for you to see the numbers.
Any clarity people can provide on this point (either in general or in specific relation to the example question I posted) would be much appreciated.
 A: The bicycle cannot be suppling only a single force against the air.   If it did it would accelerate in the opposite direction.  Instead it must also create some forces against the ground.  In a reference frame with the ground at rest, there is no displacement, so the energy/power from that force can be ignored.  In a reference frame where the ground is not at rest, there will be power associated with that force that must be considered.
In a frame where the ground is advancing, the force from the ground onto the cyclist is positive, so there is extra energy (from the ground) that can be placed into the wind.

What is unsettling me is in one frame energy is either flowing out the ground into the air or out of the air and ground into the cyclist, but in another frame the energy is flowing out of the air into the ground or out of both into the cyclist.

That's expected.  In the frame where the ground is moving forward (same direction as the relative motion of the cyclist), the cyclist is pulling energy from the motion of the ground.  In the frame where the ground is moving backward, the cyclist is putting energy into the ground.

Is it right to conclude energy is accumulating in the cyclist in when there's a crosswind like in the example?

No.  While the cyclist can convert some chemical energy into forces/kinetic energy, that does not happen in the reverse.  Any energy from ground can only go to increase the speed of the bicycle, or be delivered elsewhere (air resistance/drag).
If the cyclist is at a constant speed in some inertial frame, then the total amount of power onto the cycles from all forces (pedals, ground, air) must be zero.
In your other link, I don't understand how X->Y and and Y->X can be anything other than the negative of each other.  If the ground is doing work on the cyclist, then the cyclist is doing negative work on the ground.
A: 
Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force?

The velocity of the body on which the force is acting. More specifically, the mechanical power delivered to a system is $\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material of the system at the point of application of the force $\vec F$ on the body.
If the velocity of the material of the system at the point of application of the force is the same as the velocity of the environment at the point of application of the force, then all of the mechanical energy that leaves the environment enters the system. If the velocities differ then mechanical energy is being destroyed at the point of contact, e.g. it is converted to heat with sliding friction.

I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of

Mechanical power is frame dependent, but energy is conserved in all frames. Things like the mechanical power converted to heat from sliding friction is the same in all frames, even though the amount of mechanical power transferred varies.
A: Given a reference frame, the total power of a force field acting on a system is the sum of the dot product of each force and the velocity of the point where the force is acting
$\displaystyle P = \sum_i \mathbf{F}_i \cdot \mathbf{v}_i$,
being $\mathbf{F}_i$ lumped forces. If you deal with continuous distribution of forces per unit volume $\mathbf{f}$ in volume $V$, forces per unit stress $\mathbf{t_n}$ on surface $S$ or force per unit-length $\boldsymbol{\gamma}$ on line path $\Gamma$, summation is replaced by integration over the corresponding domains,
$\displaystyle P = \sum_i \mathbf{F}_i \cdot \mathbf{v}_i + \int_V \mathbf{f} \cdot \mathbf{v} + \int_S \mathbf{t_n} \cdot \mathbf{v} + \int_{\Gamma} \boldsymbol{\gamma} \cdot \mathbf{v} $.
Power of forces on a rigid body performing translation, not rotation. In this situation, the velocity of all the points of the body is constant, $\mathbf{v}(\mathbf{r}) = \overline{\mathbf{v}}$, and thus the power becomes
$\displaystyle P^{transl} = \left[ \sum_i \mathbf{F}_i + \int_V \mathbf{f}  + \int_S \mathbf{t_n} + \int_{\Gamma} \boldsymbol{\gamma} \right] \cdot \overline{\mathbf{v}} = \left[ \mathbf{F}^{tot,lump} + \mathbf{F}^{tot,V} + \mathbf{F}^{tot,S} +  \mathbf{F}^{tot,\Gamma} \right] \cdot \overline{\mathbf{v}} = \mathbf{F}^{tot} \cdot \overline{\mathbf{v}}$.
Kinetic energy theorem. The kinetic energy theorem states that the time derivative of the kinetic energy of a closed system equals the total power of the forces
$\dot{K} = P^{tot}$,
see https://physics.stackexchange.com/q/735204.
Kinetic energy theorem and change of reference frames. When you change the reference frame used to evaluate the position and the velocity, both the kinetic energy and the power of forces change, but the kinetic energy theorem still holds.
See https://physics.stackexchange.com/q/734777
A: 2D case the components of the force vector are
$$\vec F=\begin{bmatrix}
  F_x \\
  F_y \\
\end{bmatrix}$$
the components of the velocity vector are
$$\vec v=\begin{bmatrix}
  v_x \\
  v_y \\
\end{bmatrix}$$
where the  components  can  be taken in any coordinate  system but  the same  system for the force and the velocity vectors
from here
$$\vec F\cdot\vec v=F_x\,v_x+F_y\,v_y$$
analog 3D case
