The net force on the water at the zenith is indeed downwards.
However a net force downwards does not mean that the motion will be downwards, only the acceleration.
And the water is indeed accelerated downwards at that brief instant, which is why it continues in a circular motion, rather than flying off horizontally as it would if there were neither centripetal nor gravitational forces acting on it.
Stop the water at the zenith by stopping the bucket, and the downward acceleration will act upon it for long enough for the water to indeed move downwards - by falling out of the bucket.
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
Actually, this is rather insightful. The normal force from the ground does not quite cancel out the effect of gravity. The difference between them is precisely the centripetal force that keeps you rotating around with the Earth's surface.
Of course, you won't notice this because the centripetal force is so small compared to the gravitational force on you. The centripetal acceleration at the equator is $$a_c = \omega^2 r \approx \biggl(\frac{2\pi}{24\ \mathrm{h}}\biggr)^2\times 3959\text{ miles} = 0.034\ \frac{\mathrm{m}}{\mathrm{s}^2}$$ which is a paltry one-third of a percent of the gravitational acceleration, and at higher latitudes it is correspondingly less.