Why does the friction act on the inward direction when a car makes a turn on a level road?
Your confusion is a common one. It applies to all uniform circular motion. Consider a rock tied to a string, and you swing the rock in a circle at a uniform speed around your head.
You feel the rock pull outward on you, and this leads you to think the force is outward. This is correct. That is the force the rock exerts on you.
Another question is what you are doing to the rock. Left to itself, the rock would fly in a straight line. You are pulling on the rock. The direction of the force is along the string. You are pulling it away from a straight line toward yourself.
The two forces - you pulling on the rock, and the rock pulling on you - are equal in strength, and opposite in direction.
For a car, it isn't as easy to see two equal and opposite forces, but they do exist. But we will only consider the force exerted on the car by the road. Left to itself, say if it was sliding on ice, the car would go in a straight line. Turning the tires would not change that.
On a road, if you turn the wheels to the left, they push the car to the left. That is, there is a lot of friction from the road that prevents them from sliding. The road pushes the tires to the left to prevent sliding, and the tires push the car.
If you keep turning, you go in a circle. You see the force to the left is toward the center of the circle.
One way to answer your question:
Suppose the acceleration was not perpendicular to the "center", then $a$ would have a component in the direction of motion, which would resolve a change in the norm of $v$(speed). However, the speed was unchanged, a contradiction. Thus $a$ must be perpendicular to the direction of $v$.
(Perhaps looking for vector calculus, calculus III.)