Why didn't he consider the centripetal force in the left hand side of the equation?

He is doing right . Centripetal force is not any particular force ,any force acting in the radial direction behaves as a centripetal force,so your thinking that we should add mv²/R as centripetal force in LHS always, is wrong.

The net of all the forces acting in the radial direction is the centripetal force.The role of centripetal force can be played by any force be it gravitational, electrostatic etc.

Here ,-Nsin(phi) is behaving as a centripetal force so we are keeping it equal to ma(mass×radial acceleration). -ve sign represents that it is acting opposite to the direction of centre of curvature.

I hope you got it now?

First of all let me clear you that whenever a body moves in a circular path there is always an acceleration radially inwards which we refer to as Centripetal acceleration. And then there must be a force associated with this acceleration again radially inward. We call this force as Centripetal force. Here is the proof of the presence of a centripetal acceleration and thus a centripetal force...

https://i.imgur.com/hy1XiFp.jpg

https://i.imgur.com/Ajqxn6V.jpg

Let me make it simple for you. Consider a quarter of the circle. For example think of the top left quarter. Now the velocity vector of the body at the start of the quarter is pointing left, and at the end of this quarter it's pointing down. So there must be a force acting on the body during this passage so that the horizontal component of velocity comes to a zero starting from the value v and the vertical component of the velocity reaches the value v starting from zero. So you can see that there is a horizontal deceleration and a vertical acceleration during the passage. The net of these two gives what we call the centripetal acceleration, thus a centripetal force. I hope you have seen the mathematical proof given in those links...

Ok so, centripetal force is a basic necessity for circular motion. Think of it this way. It's not that the centripetal force exists due to circular motion but that the circular motion exists because of centripetal force. It is a physical force(meaning it arises from physical interactions, for example, when you push a box, it's a physical interaction, when two masses attract each other due to gravity, it's a physical interaction, when you stretch a rubber, it's a physical interaction.... etc etc.) acting on the body allowing for the circular motion. So whenever you encounter a circular motion you must be sure that there are some physical forces(again physical interactions) acting radially inwards that makes this motion possible... For example, tie a stone to some rope and make it rotate. You know why can the stone move in a circle? It's because of the tension in the rope acting radially inwards. Imagine what would happen if the rope suddenly vanishes? The stone will continue straight from the point when the rope disappeared. So it is the centripetal force which allows for circular motion and not vice versa... You have to understand this...

Hope it helps...

Why didn't he consider the centripetal force as a force acting on that object.

Because Centripetal force is not a force acting on the body. The Centripetal force is not necessarily a unique force in its own right. As an answer pointed out, The "centripetal force" is simply defined as the net force acting towards the centre.

If there were some more forces acting in the radial direction (for example, if your car was tied with a rope to a tree at the centre of the circle, there would be an additional force of Tension towards the centre) then the sum of the forces would be equal to Tension + the Normal reaction component. This sum would then be considered the centripetal force.

Here is another way to thing of the scenario : Write the Newton's law equation for the radial direction without thinking anything about being on a circular trajectory. Don't think anything about centripetal forces for a moment. If were on any other ordinary track, the only force on the car would be $Nsin(phi)$ which then by Newton's law is equal to $ma$. $$Nsin(phi) = ma$$

Now, because you are moving in a circular path, the "a" in the equation must be towards the centre of the circle and its magnitude is a function of the linear velocity at which you are moving in the circular path (specifically it is $frac{|v(t)|^2}{r}$ but in your case, v is constant). Now, this is something you have to accept, in circular motion, there must be an acceleration pointing to the centre with the magnitude mentioned above, otherwise the motion would not be circular anymore. Plug this into the equation and you simply get the result as : $$Nsin(phi) = frac{mv^2}{r}$$ This pretty much what the other answers said, that the sum of all forces towards the centre (there is only 1 in your case) is the centripetal force. I have taken the opposite sign convention but it doesn't affect the final result.

To answer one of your side questions, if $phi = 0$, essentially when the road is unbanked and frictionless (as per the diagram), you would never be able to make a turn (In reality, there is always friction to provide the necessary force towards the centre).

Hope this helped.

Why didn't he consider the centripetal force as a force acting on that object.

He did!

The centripetal force is not a new force - it is just a name we give to whichever force that causes the circular motion.

In his example the force that causes the inwards acceleration so that turning/circular motion is possible, is a component of the normal force. That force is then called the centripetal force, when we want to emphasise that fact, because it is causing the centripetal acceleration. But it's origin is that it is a component of the normal force.

In short: There is no new centripetal force. Centripetal force is a name given to other forces if they cause a centripetal acceleration.