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...
before attempt to turn, there was kinetic friction force only.
Where do you think you have kinetic friction? Kinetic friction is found when you have two surfaces sliding against each other. The tire-road interface is not sliding, so it is static friction, both before and during a turn.
During acceleration, the force of the tire on the road is rearward, so the reaction force of the road on the tire is forward. The only difference in the turn is that the frictional force is directed sideways instead.
If the forces between the tire and the road exceed the maximum static friction, then the tire will skid rather than roll.
Best Answer
Regarding things that are rolling such as wheels of a car, remember one key thing: kinetic friction is not about moving, but about sliding.
Even though a wheel is moving, it isn't sliding over the surface. There is no kinetic friction going on. Only static friction which holds the contact point still while it is in contact.
Since there is no kinetic friction happening when the car is driving in a circle, only static friction is left to cause the centripetal acceleration.
Now, as @JohnForkosh mentions in a comment, another way to answer your question is that the driving direction and the centripetal (radial) direction are perpendicular and thus completely seperate.
There can easily be sliding (kinetic friction) in one direction but stationarity (static friction) in the other. And this is the case here. Even if the car was sliding in the driving direction, it is still not sliding in the radial direction (it is not moving further away from the circle centre).