Pseudo forces are applied only when a non-inertial frame of reference is being used. In this case the frame of reference is the ground, which is inertial. $a=9m/s$ is the acceleration of the truck relative to the ground. $a_0$ is the acceleration of the CM of the discs relative to the ground.
It is possible to solve this problem using a non-inertial frame which moves with the lorry. Then a pseudo-force of $-ma$ must be applied to the CM of each disk. The value found for the friction force in this frame will be the same as that in the ground frame.
In the inertial frame the only unbalanced force acting on the disk is friction $f$ with the floor of the truck. There is no pseudo-force. $f$ opposes relative motion, so it is in the forward direction, the same as $a$.
The force $f$ applied at the rim is equivalent to a force $f$ applied to the CM of the disk together with a torque $fR$ about the CM.
Applying Newton's 2nd Law to the CM of the disk, $f=Ma_0$ , where $M$ is the mass of the disk. Likewise, the torque $fR$ on the disk is equal to $J\alpha$ where $J=\frac12MR^2$ is moment of inertia and $\alpha$ is angular acceleration. Eliminating $f$ we get $R\alpha=2a_0$.
The torque rotates the disk backwards relative to the truck. There is no slipping between the disk and the truck so the acceleration of the CM of the disk relative to the truck is $-R\alpha$.
Relative to the ground the acceleration of the CM of the disk is $a_0=a-R\alpha$ therefore $$a_0=\frac13 a$$ $$R\alpha=\frac23 a$$ $$f=\frac13 Ma$$