The initial premise of your question is not, in general, correct.
Consider a 1000 kg car driving at a constant speed of 20 m/s around a flat circular racetrack with a radius of 500 meters. Note the constant speed: the car will take the same time to drive around the track, 157.1 seconds, hour after hour. The only change in the car's velocity is in the direction of the velocity, not its size. The only acceleration is the centripetal acceleration, and the only horizontal force needed is the centripetal force $$a_{cp}=\frac {v^2}{r}=0.8 \frac{m}{sec^2}$$ $$F_{cp}=m \times a_{cp}=800 \ newtons$$
The force exerted by the car's engine serves only to balance the various drag forces.
This force is supplied by the friction between the tires and the pavement; pour out some oil on the track to see what happens when the required centripetal force is not present!
The only acceleration is directed exactly towards the center of the circular track; there is no tangential acceleration. The car, at some moment, is travelling North at 20 m/s, and one half-circuit later, it is travelling South at the same speed. Clearly, it has accelerated.
Assume now that the driver presses on the brake pedal in a manner that she knows will bring the car to a stop in 40 seconds. There is now a tangential acceleration, at $-0.5 \frac{m}{sec^2}$, in addition to the centripetal acceleration above, and a tangential force of 500 newtons directed towards the back of the car. The total force needed from the tires is now the resultant of these two forces: 943.4 newtons directed around 32 degrees aft of inward. Touching the brakes could throw you into a skid! Of course, as the braking changes the car's speed, the centripetal force will decrease...
You have to consider the two axes separately: lets call $x$ the horizontal plane, and $y$ the up and down (direction in which gravity acts). In the "free-body diagram" of the ball, there are also three forces to consider: 1) gravity, 2) centripetal, and 3) the tension in the string. The tension in the string needs to balance the combination of gravity and centripetal acceleration.
The centripetal acceleration itself, has nothing to do with gravity, it only has to do with the ball's circular motion. Instead of the ball spinning nearly-horizontally, you can imagine spinning it ever so slowly---and it remaining nearly vertical. Gravity hasn't changes, only the centripetal acceleration, and thus tension in the string---via force balance in the $x$ direction.
Because there is gravity in the $y$ direction, the string can never be fully horizontal, because some amount of the string tension must act in the $+y$ direction to counteract gravity. Thus, only the $y$ component of the string-tension is determined by gravity, and this occurs regardless of the centripetal force.
Best Answer
For equilibrium, the horizontal string has really to be at an angle $\theta$ as shown in the diagram.
Then resolving forces vertically on the large mass, $T=Mg$
the centripetal force is from $T\cos\theta$
and the small mass is supported by the vertical component of the tension $T\sin\theta$