You assume that the force which acts on the car is the same one that acts on the passengers, which is not correct since the car and the passengers are two different objects.
What you actually feel when the car accelerates forward is the chair pushing you forwards and you pushing the chair backwards. When there is no contact with the chair your body tends to remain in equilibrium (remains at the same velocity). As soon as your body makes contact with the chair, the chair acts with forward force on the body which you can feel.
Consider a different example which is a bit more intuitive - a passenger on roller skates in an empty wagon that accelerates at $a$ with respect to earth. The roller skates are introduced to cancel the friction force as much as possible.
The passenger and the wagon are two different objects, which means that force acting on the wagon is not the same force acting on the passenger. When there is a net force applied to the wagon, the wagon starts accelerating relative to a stationary (inertial) reference frame $A$. Since no force is applied to the passenger, the passenger remains in equilibrium as seen from $A$, which follows directly from the first Newton's law.
In the wagon (non-inertial) reference frame $B$, which moves together with the wagon in the same direction, the passenger appears to be accelerating at $-a$ (going backwards) although no force has been applied to them. This clearly violates the first Newton's law which says
An object acted on by no net external force remains in equilibrium, i.e. has a constant velocity (which my be zero) and zero acceleration.
This is why we must introduce a force that acts on the passenger in order to make the reference frame $B$ satisfy the first and second Newton's laws. Since this force actually does not exist, it is called a pseudo-force or fictitious force.
To describe this with equations
$$a_{P/A} = a_{P/B} + a_{B/A}$$
where $a_{P/A}$ is the passenger acceleration in $A$, $a_{P/B}$ is the passenger acceleration in $B$, and $a_{B/A}$ is the wagon acceleration in $A$. Since no external force acts on the passenger $a_{P/A} = 0$, the above equation becomes
$$a_{P/B} = -a_{B/A}$$
Suddenly, the passenger has some acceleration although there is no external force acting on them. Introduce a fictitious force to make the first and second Newton's laws work in $B$ and problem solved. However, this clearly violates the third Newton's law since the fictitious force has no reaction pair, i.e. there is no force that acts from the passenger to the wagon. This is why non-inertial reference frames are not suitable for the Newton's laws.
Best Answer
The total momentum the notes are talking about is the net momentum of all the particles. In whichever way the particles collide or interact with each other, they total change in momentum is 0 as the forces exerted by the particles on each other are internal forces. There are no net external forces here. If we observed this system from a non inertial reference frame, there would be a pseudo force acting on all the particles which is an external force and hence there would have been a change in the total momentum.
Yes one can see an object accelerating in an inertial reference frame, if it has some force is being exerted on it. But in the earlier example of a system of particles there were no net external forces. From any inertial reference frame, the net momentum of the particles will remain constant. But from a non inertial frame, one can see objects which don't experience any net external force with respect to inertial frame accelerate due to the pseudo force. Eg - A ball lying on the ground doesn't experience any net external force. From whichever inertial frame you observe the ball, it won't have an accelerated motion. Eg - A lift going up at a constant velocity. But from a non inertial frame , the ball will appear to be accelerated.
An object which doesn't accelerate in an inertial frame won't accelerate in any other inertial frame but will in a non inertial one due to the pseudo force.