Balancing forces is useful, but it is not enough. There aren't enough constraints on the problem to find all the forces just by setting the net force to zero. In your diagram, you could add any value to $\vec{F_h}$. As long as you add the same to $\vec{T}_c$, the net force doesn't change. So find some sum of forces that comes to zero, then add 1 Newton to both horizontal forces on the problem, and you have a new zero-net-force solution.
On the other hand, as long as the net force and net torque are both zero on a stationary rigid object, it will stay stationary. We need both because net force only tells us about the acceleration of the center of mass. If a beam starts spinning around its center, it has zero net force on it (because the center of mass isn't moving), but there was some net torque that got it spinning. So setting $F_{net} = 0$ is not enough.
As for what happened in your work, it's hard to tell. First, try to stick with well-defined notation. This will make it much easier for people to understand what you mean. Your question defines the variables $\vec{T}_c$, $\vec{F}_v$, and $\vec{F}_h$, which you didn't use. Your proposed solution uses the undefined variables "Fhx" "Fhy" and "Ft". Then you write about "theta" without defining it, either.
My best guess is that you assumed that the net force from the wall on the beam acts along that beam. That is, in the notation of the problem, you assumed
$$\frac{\vec{F}_v}{\vec{F}_h} = \frac{a}{b}$$
but there is no reason to assume this. That sort of assumption works for strings or ropes, which can support any shear, but for beams, the wall can exert forces in the horizontal and vertical directions independently. You won't know how big those forces are unless you impose both the condition that the net force on the beam is zero, and the condition that the net torque on the beam is zero.
First, some information to motivate my answer.
In circular motion, it is useful to break forces into components: the radial component $F_r$ (points towards or away from the center of the circle around which the particle is moving) and the tangential component $F_\theta$ (points tangent to the circle around which the particle is moving).
For planar motion, we can derive equations of motion for these components:
$$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$
where $(r,\theta)$ is the polar coordinate, and a dot above a variable represents a time rate of change (so, for example, $\dot r$ is just the time rate of change of $r$).
The case you are interested is circular motion, where $\dot r$ and $\ddot r$ are both $0$. Then we have
$$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$
Therefore, using Newton's second law and using the relation $\dot\theta=v/r$, we have
$$F_r=-\frac{mv^2}{r}$$
where the negative sign represents the fact that the net radial force component needs to point towards the center of the circle.
I want to point out to you a key thing in this line of reasoning. Instead of thinking of "which force is 'the centripetal force'?", instead all we need to do is determine all of the radial force components, and then we just set this net radial component equal to $mv^2/r$ towards the center of the circle. There is no single "centripetal force" in general.
So now, let's look at your question:
If we are looking at this stone moving in a vertical circle, and if we want to analyze the forces acting on the stone at the top of this circle, we need to consider the radial force components. Well, we know that the entire tension force acts towards the center of the circle, and we know that the weight of the particle acts downwards, which here is also towards the center of the circle. Therefore, based on the argument above it must be that
$$-T-mg=-\frac{mv^2}{r}$$
Now, what if we want to think about the minimum speed the stone needs to have at the top of the circle to maintain circular motion? Well, the weight $mg$ is constant for our stone, so it must be that the speed $v$ determines the tension $T$ (or you could argue it is the other way around). Based on our above equation, we have
$$v^2=\frac{(T+mg)r}{m}$$
You should convince yourself that based on this equation we can say that the minimum value of $v$ must mean that $T$ is also at its minimum value. Since the tension force cannot become a force that points away from the circle, it must be that the minimum of $T$ is $0$. Therefore, to determine the minimum velocity needed at the top of the circle for circular motion to still occur, it must be that $T=0$.
As for your second paragraph, you should now be able to see that the equation you have written is incorrect. However, that equation would be correct for looking at the bottom of the circle. But in this case tension cannot be $0$ (can you see why based on the information in this answer?).
Based on everything above, I want to specifically address your questions now
But isn't it the tension which provides the centripetal force $mv^2/r$?
Tension does always point towards the center of the circle, but this does not mean it is the only force that has a radial component. The weight of the stone also has a radial component (except for at points where the rope is horizontal and the weight is completely tangent to the circle). In general you should start with $F_r=-mv^2/r$ and then determine what $F_r$ is.
When we consider this to be zero, how come do we equate centripetal force to gravitational force?
Based on the above discussion, it should be obvious that since $T+mg=mv^2/r$ at the top of the circle, it must be true that $T=0$ means $mg=mv^2/r$. This makes sense, because now the only force acting on the stone is gravity, and this force only has a radial component.
Also, the equation $T−mg=mv^2/r$ holds for the bottom most point. How is this true when we consider mv2/r to be centripetal force whose direction is always towards the center? If this equation holds true, won't this mean that tension is 'pushing' the stone outwards when tension is considered a pulling force?
I have corrected the statement to correctly say this holds at the bottom of the circle. A better way to view this equation is $T=mg+mv^2/r$. This shows that at the bottom of the circle the tension force (still pointing in towards the center of the circle) is actually greater than just the weight of the stone.
Best Answer
If a cable is taut, then it is under tension. What you may be remembering is that, in a static situation, the total force on the cable must be zero because it is not moving. In the simplest situation of a cable being pulled at both ends, the tension forces at opposite ends are equal in magnitude and opposite in direction. This means that the total force on the cable is zero, not the tension.