The solution given in the paper is correct. The additional terms are the equivalent inertia due to the rotating wheels and terms stemming from centrifugal forces. As your chosen reference frame is not an inertial reference frame, you have include fictitious forces that you wrongfully did not consider in your derivation.
I have only taken a quick glance at the paper and the derivation but I think I can come up with a more concise and physically transparent explanation making use of classic Newton-Euler in a moving reference frame and corresponding fictitious forces. It allows you to understand more precisely where all the terms come from but I would not advise you to determine the equations of motion this way as this is very error-prone. In the following section I will use the superscript indices for indicating the point of reference, the subscript indices for indicating the part ($w$ = wheel, $c$ = cart) as well as the rotation axis ($\perp$ is rotation axis of the particular part around the axis perpendicular to the plane in which the vehicle is moving, $\parallel$ on the other hand marks the axis of rotation of the wheels).
Neglecting the rotating wheels
In the first step we will neglect the rotating wheels and just consider them as sliding across the floor. The rotation will introduce an additional inertia that we will consider in a next step.
From the free body diagram (figure 3) we can see that there are two axial forces $F_{wL}$ and $F_{wR}$ that are not convenient for us, thus we decide to apply the principle of angular momentum in point (A) where they vanish. This point is neither the center of mass nor a fixed point in the global coordinate system, it is moving relative to it. This means we have to use the general form of the principle of angular momentum
$$I^{(A)} \dot{\vec \omega} - m \left( \vec \omega \times \vec r^{(A-CM)}\right) \times \vec u^{(A)} = \sum\limits_i \vec M_i^{(A)}$$
which can also be written - in case of a two-dimensional system (where there is only one axis of rotation and one angular velocity instead of an entire vector) and further taking into consideration that the axial forces of the wheels have no contribution to the torque - as
$$I^{(A)} \dot{\omega} + m_c \, \omega \, d \, v = L \left( F_{uR} - F_{uL} \right) = \frac{L}{R} \left( \tau_{R} - \tau_{L} \right).$$
The moment of inertia around this point can be calculated using the moment of inertia around the corresponding center of mass and the parallel axis theorem (Steiner's theorem)
$$I^{(A)} = I_c^{(CM)} + \underbrace{m_c d^2}_{\text{Steiner cart}} + 2 I_{w_\perp}^{(CM)} + \underbrace{2 m_w L^2}_{\text{Steiner wheel}}.$$
You can already see that your equation is missing the centrifugal term due to the rotation of the reference point A.
Let's also use the coordinate system of the moving cart for Newton's second law. The chosen reference system is no proper inertial reference frame: If you rotate the vehicle a certain force will be felt that impacts the motion of the vehicle and must be considered in Newton's second law as fictitious forces. The second law for a rotating reference system actually takes following form
$$m \dot{\vec v}^{(CM)} = \underbrace{- m \, \frac{\vec \omega}{d t} \times \vec r^{(a-CM)}}_{\text{Euler force}} \underbrace{- 2 m \, \vec \omega \times \vec v^{(CM)}}_{\text{Coriolis force}} \underbrace{- m \, \vec \omega \times \left( \vec \omega \times \vec r^{(a-CM)} \right)}_{\text{Centrifugal force}} + \sum\limits_i \vec F_i.$$
For our 2D system the Euler force and the Coriolis force vanish due to symmetry reasons (at least in the relevant direction) but nonetheless we will have to include the centrifugal force that is felt whenever the vehicle is rotating $F_{cent} = m_c \, d \, \omega^2$
$$m \dot{v} - m_c \, d \, \omega^2 = F_{uR} + F_{uL} = \frac{1}{R} \left( \tau_R + \tau_L \right).$$
Including the rotation of the wheels
So far we have neglected the rotating wheels and treated them as if they were just sliding across the floor. If we consider rotating wheels we have to account for the fact that the rotating wheels pose a resistance to every kind of change of motion, they increase inertia. We have already considered the inertia of the wheels moving with the vehicle but we have to consider the relative motion of the wheels now. Let's introduce a reduced mass $m_{red}$ with the corresponding linear velocity $v$ that requires the same amount of energy as accelerating a wheel with rotational inertia $I_{w_{\parallel}}$ and rotational velocity of a spinning wheel $\dot{\varphi}$
$$\frac{1}{2} I_{w_{\parallel}} \dot{\varphi}^2 = \frac{1}{2} m_{red} v^2.$$
With $R \dot{\varphi} = v$ for a single rotating wheel we can instantly see that we would have to add two times the fictional reduced mass of an accelerating wheel
$$m_{red} = \frac{I_{w_{\parallel}}}{R^2}.$$
Similarly if the two wheels spin in opposite directions this introduces an additional moment of inertia that has to be overcome
$$\frac{1}{2} I_{w_{\parallel}} \dot{\varphi}^2 = \frac{1}{2} I_{red} \omega^2.$$
If only one wheel is spinning the rotation caused by this wheel must be equivalent to the rate of rotation $\dot{\varphi} \, R = \omega \, L$ and therefore
$$I_{red} = \frac{L^2 \, I_{w_{\parallel}}}{R^2}.$$
These observations can be extended to arbitrary configurations with $v = \frac{R (\dot{\varphi}_{w_R} + \dot{\varphi}_{w_L})}{2}$ and $\omega = \frac{R (\dot{\varphi}_{w_R} - \dot{\varphi}_{w_L})}{2 L}$.
Full equations of motion
Combining the equations of motion for the vehicle and the rotating wheels results in
$$\left(m + \frac{2 I_{w_\parallel}}{R^2} \right) \dot{v} - m_c \, d \, \omega^2 = \frac{1}{R} \left( \tau_R + \tau_L \right),$$
$$\left(I_c^{(CM)} + m_c d^2 + 2 I_{w_\perp}^{(CM)} + 2 m_w L^2 + \frac{2 L^2 I_{w_\parallel}}{R^2} \right) \dot{\omega} + m_c \, \omega \, d \, v = \frac{L}{R} \left( \tau_{R} - \tau_{L} \right).$$
