Because I have a closed loop, should there be some sort of "back action" from the tension in one leg of the loop on that on the other leg?
Cool question. About the "back action." Imagine, instead of a smooth cylinder, a cylinder with a thin, rectangular bump directly opposite the point that the load hangs beneath (on the top part of the cylinder.) Also, instead of a circular rope which stretches all the way around, picture a regular, two-ended rope which stretches around the underside of the cylinder (that is, the load side), but where each end of it is joined to one side of the bump, on the top.
In this case, there cannot be any back action. But this is the same as the case you describe. Let us illustrate. First, we can cut the bump from the cylinder -- nothing will change, as there is no net force upwards or downwards acting on the bump. Second, we can replace the bump with a piece of rope, and voila -- this is just a rope stretched all the way around.
Since tension only comes from forces acting parallel to the direction that the rope is stretched, and the normal force from the cylinder is always perpendicular, there can be no contribution to the tension from the cylinder.
This means the tension in the circular rope, due to the load, is the same as the tension in two straight, fixed ropes carrying that load.
Also, it should be obvious that the sling exerts force on the tree, but if I only look at tension, which runs in the direction of the rope, I can't really tell where that force is coming from.
The rope now pulls down on the limb a weight equal to the weight of the load (if we take the mass of the rope to be negligible.) The pressure on the cylinder does not come from the tension in the rope.
Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall.
Let's see why. Let's decompose the weight of the stick into a component along the stick, and one perpendicular to it. The latter component creates a torque around the point $O$. Let's calculate this torque.
$$\tau = g\int_A^B R \ \rho (R) \ \text dR \tag{1},$$
where $R$ is the distance from the point $O$, $\rho$ is the density of the stick per unit length, $A$ and $B$ are the extremities of the stick.
Now let me express $R = R_0 + r$ where $R_0$ is a point whose meaning will appear below. We get
$$\tau = g R_0\int_{R_A - R_0}^{R_B - R_0} \rho (R_0 + r) \ \text dr \ + g \int_{R_A - R_0}^{R_B - R_0}r \ \rho (R_0 + r) \ \text dr. \tag{2}$$
Well, the 1st integral gives
$$\tau = g R_0 M = GR_0 \tag{3}$$
where $G$ is the weight of the stick, and this is the final result if we choose the point $R_0$ in such a way that the 2nd integral in (2) be zero. The point $R_0$ for which the 2nd integral in (2) vanishes is the center-of-mass. As to the result (3), it says that the torque imposed by the weight of the stick is equal to the distance to the center-of-mass, $R_0$, times the weight of the stick as if it were concentrated, all of it, in the center-of-mass.
Two simple examples If the stick has uniform density, then the 2nd integral yields the expression $g\ \rho \frac { (R_B - R_0)^2 - (R_A - R_0)^2}{2}$ which is clearly zero for $R_0$ chosen in the middle of the stick. But if, say, the lower third of the stick is has four times the linear density of the rest, the 2nd integral is zero for $R_0$ chosen at 1/3 height, i.e. $R_A - R_0 = -L/3$, where L is the stick length, and $R_B - R_0 = 2L/3$. Indeed, $g\frac {1}{2}[-4\rho \frac {L^2}{9} + \rho \frac {4L^2}{9}]$.
Best Answer
Realizing that accepting my previous answer would be based on you accepting and understanding the definition and properties of couples, the following is intended to help in that regard.
Definition of a couple:
A moment created by two equal, opposite and parallel non-collinear forces.
Fig 1 below is an example of a couple. The net force is zero. Each force causes a clockwise moment about the point in the center equal to $\frac {Fd}{2}$ for a total moment of $Fd$. Note that the magnitude of the couple is the same if larger forces are closer together or smaller forces are further apart. For this reason showing the forces is unnecessary and the symbol to the right of the example is often used.
1. It is equivalent to a single moment vector.
The symbol of the couple reflects this property. For a couple formed by forces in the x-y, by the right hand rule a counter clockwise couple vector is in the positive z direction and is typically indicated as positive.
2. It induces the potential for rotation without translation.
This is evident from Fig 1. There is no net force.
3. In order to achieve equilibrium for the entire beam, a couple can only be counteracted by another equal and opposite couple.
To illustrate this property, consider the simply supported beam of Fig 2A. A single force couple is applied to the center of the beam. The static equilibrium requirements are the sum of the forces and sum of the moments are zero.
Summing the forces we obtain $R_{A}=-R_{B}$. The reaction forces are equal, opposite and parallel to each other. Therefore they form a couple.
Summing the moments we get $R_{A}=-F/2$ and $R_{B}= +F/2$. Therefore the couple formed by the reactions is counter clockwise, countering the applied couple.
4. The moment vector of the couple can be moved to any location without affecting static equilibrium requirements.
Note that the location of the applied couple (point C) in Fig 2 had no influence on the sum of the forces for equilibrium. No matter where it is, the reactions will be equal and opposite.
Likewise, the location of C will have no influence on the sum of the moments for equilibrium. For example, in Fig 3 the location of the couple, point C, has been moved so that it is now $L/4$ to the right of A instead of in the center. The sum of the moments about A remains zero.
Hope this helps.