To start off, it seems like you are thinking about uniform circular motion, as you are fixated on the centripetal acceleration and are not mentioning anything about tangential acceleration. Therefore, for now let's assume we are talking purely about uniform circular motion.
For circular motion to happen we need a centripetal force or acceleration i.e. a force perpendicular to the direction of motion at all instants. So without any perpendicular component of force circular motion will not occur.
Does the above statements mean that the circular path is a resultant path of the given velocity and the velocity due to the centripetal acceleration?
If by "velocity due to the centripetal acceleration" you mean $\text d\mathbf v=\mathbf a\,\text dt$, then yes; this is just applying the definition of acceleration $\mathbf a=\text d\mathbf v/\text dt$ to the velocity:
$$\mathbf v(t+\text dt)=\mathbf v(t)+\mathbf a\,\text dt$$
This is true for all motion, not just circular motion.
Why doesn't the particle gain any velocity in the centripetal direction although it has some acceleration in that direction? A horizontally projected body gains some velocity in the direction of mg, so a body in a circular motion should also gain some velocity in the centripetal direction.
Something to remember is that the centripetal direction changes as the particle goes around the circle. The particle does gain velocity in the centripetal direction, but since the velocity was along circular path the instant before, once the object does pick up this velocity component, that component is no longer completely centripetal. The velocity does change to be more "aligned with" the acceleration vector, but since the acceleration vector is always changing directions the velocity vector will be constantly trying to align with different directions, and so we get the circular motion you describe.
This is different from the projectile case you give, where the vertical direction is a constant direction.
I am adding a picture to show what I think about circular motion. Imagine the picture as a magnified image of very small distance.
Here $v′$ shows the velocity due to centripetal acceleration and the dot at the center shows the cause of centripetal acceleration and dotted lines represent the path it would have taken with no force on it.
There are two issues with your diagram:
1)It looks like your centripetal acceleration is too large
2)It looks like your $\Delta t$ is too big
In order for circular motion to occur, the centripetal acceleration has to be exactly equal to $v^2/r$. It is not sufficient for the acceleration to just have a component perpendicular to the velocity at all points in time. From your diagram, it is obvious that $v'=a\text dt$ is too large.
To explore this more, let's numerically solve the differential equations with a larger than required centripetal acceleration. For planar motion in polar coordinates, we normally have the differential equations
Since we are assuming no tangential forces, let's set $a_\theta=0$. Now, if we correctly made $a_r=-r\dot\theta^2$, then we would be left with $\ddot r=0$, which would give us our uniform circular motion for $\dot r(t=0)=0$. However, let's impose a centripetal acceleration to be just a little larger than this (still dependent on the velocity) $a_c=-1.1r\dot\theta^2$, so we have the differential equation $\ddot r+0.1\cdot r\dot\theta^2=0$.
Solving the system of differential equations
with initial conditions (dropping units) $r(0)=1$, $\dot r(0)=0$, $\theta(0)=0$, $\dot\theta(0)=1$, we get the trajectory
And we see that we get spiraling inward
Similarly, for $a_c=-.9\,r\dot\theta^2$, we get a trajectory that goes away from the origin
Of course, this is not exactly like your diagram since now the acceleration is not exactly perpendicular to the velocity, but if we had a spiral trajectory where the acceleration was always perpendicular to the velocity then we would have to abandon the assumption of a non-tangential acceleration.
This does however relate to the second issue in your diagram; you are only applying a perpendicular acceleration at set times rather over the entire trajectory. Now, I know we can always approximate the change in velocity as $\mathbf v(t+\Delta t)\approx\mathbf v(t)+a\Delta t$, but if $\Delta t$ is too large, then you are not going to get the correct trajectory.