I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand.
Note: it's a little difficult to do this fairly with your hand. When I tried it, I had a tendency to slightly adjust the plane of motion of the mass, so that it oscillated slightly. This was particularly bad when trying the second situation. If I didn't do that, the sting would hit one of my knuckles every time it passed. That's another indication that the plane of revolution is actually below my hand.
Your force diagrams are qualitatively correct. Gravity points down towards the floor, and the tension points along the string at some angle. It's easiest to break the tension into a vertical component (which will either add to or subtract from gravity), and a radial component, which lies in the plane of the ball's orbit and provides the centripetal force.
To be concrete, take $\theta$ to be the angle between the string and the vertical. The string hanging under just gravity means $\theta = 0$, and the ball orbiting in the horizontal plane is $\theta = 90^\circ$. You want the vertical component to cancel out gravity, so $w = T\cos{\theta}$. As you increase $\theta$, $\cos{\theta}$ decreases, so you have to increase $T$ to keep the vertical component balanced. The centripetal force is $F_c = T\sin{\theta}$, which increases both because you increased $\theta$ and because you increased $T$. The ball will then need to move faster to account for the increased centripetal force. I'll let you work out the actual details, but you should get that the ball will need to move infinitely fast to get to $\theta = 90^\circ$.
An illustration of the wobble effect I noticed is when cowboys attempt to lasso animals. You can watch this video starting at 1:20 to see this.
Assume that the point mass, $m$ has two tiny thrusters, mounted so as to exert purely tangential force in the plane of the circular motion, one clockwise, and the other counter-clockwise.
The magnitude of the constant velocity of the mass is $v$, and the radius of the circle is $r$.
Measure the position of the point mass in the standard Cartesian coordinate way: angles are measured from the positive X-axis, counter-clockwise positive.
At the point where the mass is at a position angle $\theta$. the total radial force inward on the mass, $F_R$ is given by the centripetal force equation:$$F_R=\frac{mv^2}{r}$$
There are two forces that supply this radial force: the tension, $T$ in the string, and the inward radial component of the force of gravity:$$F_{G-R}=mg\sin(\theta)$$So:$$\frac{mv^2}{r}=T+mg\sin(\theta)$$and:$$T=\frac{mv^2}{r}-mg\sin(\theta)$$Note that this implies that:$$v >=\sqrt{rg}$$ or the string tension will become negative near the top of the circle, an impossibility.
The conditions of the question also require that at all times the net tangential force, $F_T$, be zero. The tangential component of the force of gravity, $F_{G-T}$ is given by:$$F_{G-T}=-mg\cos(\theta)$$where a positive force implies counter-clockwise force. The thrusters are needed to supply the exact opposite force to the mass at all times.
Best Answer
The centripetal force is not a "separate" force. I think it's best not to think of centripetal forces, but just centripetal acceleration. An object with circular motion means that net sum of all the forces acting on the object results in circular motion... meaning the net acceleration towards the center of the circle is $\dfrac{v^2}{r}$
In your situation there are two forces acting on the ball. The tension in the rope and gravity. (there's no extra centripetal force).
$\Sigma F_{towards center} = m_{ball}a_{towardscenter} => T = m_{ball}\dfrac{v^2}{r}$
So gravity does not play a role here because gravity acts downward, and the direction towards the center of the circle is to the left.
Suppose the ball was at an angle of 45 degrees to the right of the upward direction. Then you'd have to consider the tension in the rope and the component of gravity acting towards the center. Specifically you'd get $T+m_{ball}gcos(45) = m_{ball}\dfrac{v^2}{r}$
But anyway, for your question $T = m_{ball}\dfrac{v^2}{r}$