When the string becomes slack, the centripetal force disappears. So, the particle just undergoes normal parabolic projectile motion.
To get the exact motion, using the initial velocity and angle, find the parabola of its motion. Then see when it again cuts the circle of the original motion. At this point, it will re-enter circular motion keeping the tangential component of its velocity.
Note that string becomes slack when your equations say $T\leq0$
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
What is the configuration needed for a circular-motion?
The answer is, there must be an inward(towards the center) force perpendicular to the instantaneous velocity. The minimum velocity required for the bob initially to loop the whole loop is $\sqrt{5gR}$. In this situation, the tension at the topmost point becomes zero, but since there is velocity, which is $\sqrt{Rg}$ & there is inward force, which is the weight $mg$; this matches up the configuration of circular-motion & hence continues to complete the circle.
Now, if the initial velocity is less than $\sqrt{5gR}$, then the bob will have attained zero velocity before reaching the top. Now, using this velocity, we can find $h_1 \; , \; h_2$, the heights where the tension becomes zero & the height where the tangential instantaneous velocity of the bob becomes zero respectively. Using kinematics equation, we find $$h_1 = \frac{v_\text{initial}^2 + Rg}{3g} \\ h_2 = \frac{v_\text{initial}^2}{2g}$$.
Now, the case you mentioned can be described as $h_1 \lt h_2$ i.e. the tension becomes zero before velocity becomes & this implies certain velocity does exist when tension becomes zero.
The gravitational force never does point inwards i.e. towards the center of the circular trajectory. The component $mg\sin \theta$ opposes the tangential velocity & remains collinear with the instantaneous velocity vector but in the opposite direction; $mg\cos \theta$, on the other hand, points outward i.e. away from the center just opposite to the centripetal force i.e. tension. You can see in the pic$^1$ how the components of gravitational force are specified; no one is acting inwards i.e. towards the center of the circle.
When tension becomes zero, velocity still exists & $mg\cos\theta$ remains perpendicular to it & is pointing inward to some point; hence it matches the configuration of circular-motion. The bob thus breaks its old circular trajectory & undergoes circular motion around a new loop where the centripetal force is $mg\cos\theta$. But it can't undergo a pure circular motion as the bob is retarded by the opposing force $mg\sin\theta$ & thus it follows a parabolic trajectory.
So, in a nut-shell, it is gravity which lets the bob to move away from the trajectory when tension becomes zero.
$^1$ Courtsey: Understanding Physics vol.I by D.C.Pandey.