How does centripetal force changes the direction of velocity? Isn't it a violation to Newton's second law of motion? Newton's second law is a vector law.
When when we resolve it in component form along the x, y and z axes we can conclude that force changes only the component of velocity along it, for example if the only force is along x axis then only the velocity along x changes but not other two. So why, in uniform circular motion, does the centripetal force changes the direction of velocity even though it is perpendicular to velocity?
My reasoning is that at any instant say at $t = 0$ the force is along radius and perpendicular to velocity, at $t = dt$ the velocity perpendicular to force is unchanged both in magnitude and direction but a new velocity is gained $dv$ in $dt$ time which is along the radius and now the resultant velocity has the same magnitude as before approximately but a different direction.
Is my reasoning correct or is there some a other explanation?
 A: 
force changes only the component of velocity along it

Yes, indeed. So when the force is perpendicular, then a perpendicular component is added to the velocity. This original velocity vector and this new added perpendicular component then combine into a new velocity that is slightly tilted towards the force.
So, it should not be a surprise that a perpendicular force causes the velocity vector to turn.
But what indeed might be confusing is that only turning happens in the case of uniform circular motion. No magnitude change. This might seem counterintuitive since the original velocity vector combined with this newly added perpendicular component should combine into a longer vector.
The trick here which beats our intuition is that we are dealing with an infinitely small new perpendicular component which is only appearing for an infinitely short duration. It only "has time" to cause turning, but "no time" to also cause a magnitude increase.
In the next instant the force has namely turned to still be perpendicular to this new turned velocity vector. Again it causes turning  - and again without a magnitude change since it again only lasts for an infinitely short time. Repeat this constantly, and you have a circular motion with constant speed. Your own description in the last half of your question is correct.
A: One can always express the position of some object with respect to some origin as a vector in as $\mathbf r = r \hat r$ where $\mathbf r$ is the object's position vector, $r$ is the magnitude of that position vector, and $\hat r$ is the unit vector parallel to $\mathbf r$. Differentiating with respect to time yields $\mathbf v \equiv \dot{\mathbf r} = \dot r \hat r + r \dot{\hat r}$.
If the radial distance is constant, $\dot r$ is identically zero. But what about $\dot{\hat r}$? This is a unit vector, which is a special case of a constant length vector. Consider a vector $\mathbf x$ whose length is constant with respect to time: $||\mathbf x||^2 = \mathbf x \cdot \mathbf x= \text{const}$. Differentiating with respect to time yields $\mathbf x \cdot \dot{\mathbf x} = 0$. In other words, the time derivative of a constant length vector is either zero or is normal to that constant length vector. A unit vector is obviously a special case of a constant length vector.
A: 
Newton's second law is a vector law. When when we resolve it in component form along the x, y and z axes we can conclude that force changes only the component of velocity along it, for example if the only force is along x axis then only the velocity along x changes but not other two. So why, in uniform circular motion, does the centripetal force changes the direction of velocity even though it is perpendicular to velocity?

Right, here's the problem. Force is a geometric quantity, a force vector is literally pointing to somewhere in space, it exists without your decomposition into components.
So, suppose I had a force vector decomposed in the cartesian basis as $]\vec{F}= F_x \hat{i} + F_y \hat{j}$, while it true that the acceleration component along x axis is related to $F_x$ and along y is related to $F_y$. The actual direction which a particle is accelerated in is $ \tan^{-1} \frac{F_y}{F_x}$ with respect to x axis.
For the centripetal force equation, to derive it we decompose the force into the 'local' basis of the whatever curve the particle moves about. I say local because, the 'unit basis' of the curve varies from point to point (I.e: the unit normal and unit tangent vectors).
For more information, see Irodov Fundamental Laws of Mechanics (It's an old book but covers this topic well at the beginning)
A: 
So why, in uniform circular motion, does the centripetal force change the direction of velocity even though it is perpendicular to velocity?

It's the very fact that the force is perpendicular to velocity, which is responsible for the change in direction of the velocity. The speed (the size of the velocity) is not changed though. At first sight, you might think that because there is no force in the direction of motion this motion can't be changed. Nothing is further from the truth though.
For a circular moving object, the length of the velocity vector (the speed of the object) doesn't change. The direction does. This is "the opposite" of the force being parallel to velocity in which case there is no change in the direction of the velocity but only in the length, i.e. the speed.
The centripetal force does make the individual components of the velocity vary. These two individual components are the speeds in two independent (non-parallel) directions which are usually taken to be perpendicular. The speed in each of these two directions changes constantly due to the centripetal force. But when you calculate the square root of the sum of the squared speeds (which is the length of the velocity vector, i.e., the total speed) the result is always the same.
A: 
So why, in uniform circular motion, does the centripetal force changes the direction of velocity even though it is perpendicular to velocity?

This happens for projectiles too. Think about the object at the height of the trajectory: the force is perpendicular to the velocity, and at that single point the velocity is changing in direction but not in magnitude.
I'm not sure why you think circular motion breaks this idea that the force component and acceleration component are in the same direction. Think of the motion at the top of the circle. It is completely horizontal, and the centripetal force is completely vertical. The velocity starts changing in the vertical direction and not at all in the horizontal direction at this point, which is consistent with what you know: force components only change the velocity components in that direction.
A: When I think of a satellite orbiting the Earth at a constant speed, I also think about the fact that its velocity is constantly changing, because its direction is changing all the time. It's direction is always changing, moving towards the center of the Earth. Accelerating towards the center of the Earth and obeying Newton's Second Law of Motion and Law of Gravitation: $F = ma = g{\cdot}M{\cdot}m/r$ squared. This acceleration and change in direction is the same direction of the only force acting upon the satellite (i.e. the gravitational force along the radius of the orbit). It makes most sense to me, if I tell myself that the satellite actually is always falling to earth, but that the horizon just keeps disappearing out of view before it can ever get there. So it never lands, it just theoretically orbits forever. Hope this helps.
A: As one can see from the image below the net force ($\vec F_{net}$), therefore the centripetal acceleration, is in the same direction with the change in velocity ($\vec \Delta v$).

A: The second law is a relation between force and acceleration. In the case of linear acceleration it is intuitive that force causes the acceleration.
For circular motion, there are some cases, as a charge moving in a constant magnetic field, or an asteroid path being defleted by the gravity attraction of a planet, where the same principle applies. Force causes acceleration.
But in most of the cases (of circular motion), it is better to think of the change of the direction of the velocity (the centripetal acceleration) causing the centripetal force. That force is the result of some type of constraint.
One example is a dumbell rotating in the outer space around its COM. There is no external forces and it simply keeps rotating. A strain gage installed in the rod between the masses will show a uniaxial stress, indicating that there is a force in the direction of the rod. And it is a centripetal force, because the masses would separate if their constraint to the rod were removed.
The same for hammer throw. The athlete must provide an angular acceleration for the ball get the desired angular velocity. As a consequence of the increasing centripetal acceleration, the centripetal force necessary to hold the ball increases. When it is released, the ball follows its tangential velocity.
