Yes, tension only affects the direction of the particle's velocity. This is because it is always perpendicular to the velocity, and because work is actually the dot product of force and displacement:
$$
W = F \cdot s = |F| \times |s| \times \cos(\theta)
$$
, a force perpendicular to the displacement does no work
The tension on the string will have an equal and opposite "reaction force" from the mass.
Note that the tension $T_{top}$ at the top is in general not the same as the tension $T_{bottom}$ at the bottom of the loop. Using the same symbol $T$ either implies that the tension is the same (which means something must be done to change the speed of the particle as it goes around), or it is just messy. Since the point in the middle is "fixed" per your statement of the problem, it must be that the particle is undergoing free motion, and we need to account for the change in velocity.
You can start with the velocity at the top of the orbit - this must be sufficient to offset the force of gravity, but tension in the string could be zero. This puts a lower bound on the velocity, and
$$\frac{m\ v_{top}^2}{r} > m\ g\\
v > \sqrt{r\ g}$$
Now the velocity at the bottom will be greater because the potential energy from the top is converted to kinetic energy at the bottom:
$$\frac12mv_{top}^2 + mgh = \frac12mv_{bottom}^2$$
And the tension at the bottom must be sufficient to account for the additional velocity as well as the force of gravity - which is now pointing outwards.
The total tension at the bottom has to be
$$T = mg + \frac{mv_{bottom}^2}{r}\\
=mg + \frac{mv_{top}^2 + 2mgh}{r}$$
I am still not completely clear what is intended with the "reaction force" on the mass - but it is reasonable to say that it is the force of the string on the mass, which is equal to the tension and is given by the above expression.
Best Answer
It's easiest to see what is happening if one looks at the object from the perspective of an outside observer:
Nothing. Nothing keeps the mass from falling down. If you really think about it, 100% of the times where the object reaches the highest point, it immediately begins accelerating downward (falling). The only thing that's obscuring the fall is that the object has a horizontal velocity at that time, so it's path isn't straight down: it's down and sideways. But it's absolutely falling.
Really the only thing that stop it from "falling" in any sense of the word is that the rope/ring "catches" it on the bottom half of the swing. But at the moment at the top of the ring, it really is just "falling."
When talking about reactions, remember that the equal and opposite reaction is always between the same two objects, but the roles of subject and object are switched. So if you talk about a force applied by the ring to the mass, the reaction force must be an equal and opposite force applied by the mass to the ring. No other force can be called its "reaction." It cannot be weight because weight is not a force applied by the mass to the ring.
It is very easy to confuse two concepts. Reaction forces are always equal and opposite, and the subject and object switch roles. Weight might balance the force of the ring on the mass, but that is not the same as a reaction force.
In your specific example, at the top of the ring you stated that the force on the mass by the ring is at a minimum. This minimum is at the point when the object experiences "weightlessness" or "freefall" where there are no (net) forces other than gravity acting on it. So in this very specific example of the mass at the top of the ring, the force of the ring on the mass is 0.
Now, with all of that inertial thinking out of the way, we can talk about the behavior in a rotating frame which follows the mass. In this rotating frame, the motion of the mass is 0 (sort of by definition: we defined the rotating frame such that, in that frame, it isn't moving). There are no new forces when we think about it this way. There is still gravity, a force from the rope/ring on the mass, and its equal and opposite force of the mass on the rope/ring. Changing reference frames does not change the true forces in the system.
The next few paragraphs are more math, just pinning down the equations of motion. For an intuitive understanding, they can be skipped, and one can pick up after them.
However, in a rotating frame, the equations of motion are different than they are in an inertial frame. There is a centrifugal acceleration. Instead of the equations of motion being $\Sigma F=ma$, they are almost $\Sigma F=ma + (\Omega\times(\Omega\times r))+(2\Omega\times v_r)$ where $\Omega$ is the angular rate of the frame (w.r.t the inertia frame), and $v_r$ is the velocity of the object in the rotating frame. The first of those terms is the centrifugal term, which you are thinking about in this problem, and the second is the Coriolis term which shows up in other problems.
I switched into vector notation there to line up with the variables used on Wikiepdia's page on rotating frames. They work in all cases. In this specific case, the motion is in one 2-d plane, and the angular rate vector is at right angles to that. In that special case, the cross products ($\times$) are simple. Let $\Omega=\omega\hat z$, where $\omega$ is a scalar (which is probably what you're used to). $(\Omega\times(\Omega\times r))$, the centrifugal term, computes to $\omega^2 |r|$ in the radial direction, $\hat r$ (the intuitive direction for the centrifugal term). The Coriolis term, $(2\Omega\times v_r)$ becomes $2\omega v_r$ at right angles to the velocity. And in your particular case, there's no velocity for this particular object in this particular frame ($v_r=0$), so we'll ignore that term. Thus we get $\Sigma F= ma + \frac{\omega^2} {|r|}\hat r$
If you skipped over the math, the takeaway from the above paragraphs is that in a rotating frame, the equation of motion is not $F=ma$. It's $F=ma + C$, where $C$ is the above expressions capturing centripetal and Coriolis accelerations. And one should note that these are accelerations not forces. Sometimes we will hand-wave them into forces, simply by multiplying in the mass and dragging them to the left side of the equation. But they are not forces. They are pseudo-forces. The difference is that every force has an equal and opposite reaction force, but that is not true for pseudo-forces. Every time someone speaks of a centifugal forces, remind yourself that it's actually a pseudo-force.
Now in the case of your example, the centripetal force from the ring is 0. Gravity is also centripetal at this one particular moment in time. These are the only true forces on the object. If we were thinking in inertial terms, we'd say $\Sigma F=ma$, and since the sum of the forces is $-mg$, acceleration must be $a=-g$. But since we're in a rotating frame, where' there's $\Sigma F=ma + C$, we have $-mg=ma + C$ where $C$ is the centrifugal acceleration associated with the rotating frame. Since in our particular example the object is not accelerating in the rotating frame, $a=0$, we get $-mg=C$. The force of graity at the top of the circle is exactly the same magnitude as the centrifugal acceleration , just in the opposite direction.