The two-wheel model shown in the question is good start for determining the loading on the tires. But this would only result with the total load on the front tires, and the total load on the back tires, without any further details on how these loads are distributed left-to-right. To get there, you also need to consider the height of the center of mass relative to the wheel axles.
In any case, I will proceed with the two-wheel model below.
The laws of motion provide with a balance of forces and well as a balance of torques for you to consider. Maybe doing a free body diagram first, might help. I hope you are familiar with high school physics, trigonometry, and the concept of vectors.
Here centerline of the front wheels is at A, the centerline of the rear wheels at B and the center of mass at C. The side forces at the rear are $\vec{F}_B$ directed towards the center of rotation O, the side forces at the front are $\vec{F}_A$ also directed towards O, driving force on the front is $\vec{P}_A$ (assume front wheel drive car), and the centrifugal force $\vec{F}_C$ is located at C pointing away from O.
Notice also that the radius of turn $R$ is located where the rear wheels are, describing an arc with radius equal to $$R = L \cot \delta$$ where $L$ is the distance $\overline{AB}$.
The rotational speed of the vehicle $$\Omega = \frac{v_B}{R} = \frac{v_B^2 \tan \delta}{L}$$ where $v_B$ is the speed on the rear wheels. Also the radius of the center of mass is $$R_C = \sqrt{R^2 + c^2} = \sqrt{\left( c^2 + L^2 \cot^2 \delta\right)} $$
The arc radius of the front wheels is needed also $R_A = \sqrt{L^2+R^2}$
So the centrifugal force magnitude is $$F_C = m \Omega^2 R_C = m \frac{v_B^2 \tan \delta}{L} \sqrt{\left( 1 + \frac{c^2 \tan^2 \delta}{L^2}\right)}$$
But the car might also accelerate or decelerate changing its rear-wheel speed with acceleration $a_B$ causing rotational acceleration $$\dot \Omega = \frac{a_B}{R}$$ and inertial force on the center of mass $$P_C = m \dot \Omega R_c = m a_B \sqrt{ \left( 1+ \frac{c^2 \tan^2 \delta}{L^2} \right)}$$
Additionally, the change in rotation requires a net torque $M_C$ to be applied to the center of mass that equals to $$M_C = I_C \dot \Omega = I_C \frac{a_B \tan \delta}{L}$$ where $I_C$ is the mass moment of inertia of the car about the center of mass. This can be estimated with $I_C \approx m c (L-c)$ by approximating the car as a lumped mass of $\tfrac{L-c}{L} m$ on the rear wheels and $\tfrac{c}{L}m$ on the front wheels.
And now for the balance of forces and torques (about the center of mass)
$$\begin{aligned}
F_B + \tfrac{R}{R_A} F_A + \tfrac{c}{R_C} P_A - (m \dot \Omega R_C) \tfrac{c}{R_C} - (m \Omega^2 R_C) \tfrac{R}{R_A} & = 0 \\
P_A \tfrac{R}{R_C} - F_A \tfrac{L}{R_A} - (m \dot \Omega R_C) \tfrac{R}{R_C} + (m \Omega^2 R_C) \tfrac{L}{R_A} & = 0 \\
c \left( F_B \right) - (L-c) \left( F_A \tfrac{R}{R_A} + P_A \tfrac{L}{R_A} \right) - I_C \dot \Omega & = 0
\end{aligned}$$
Where front-wheel forces act along the $\delta = \tan^{-1} \left( \tfrac{L}{R} \right)$ angle and center-of-mass forces act along the $\delta_c = \tan^{-1} \left( \tfrac{c}{R} \right)$ angle. The above is to be solved for $F_A$, $F_B$ and $P_A$ given a specific rotational acceleration $\dot \Omega$.
The results are very complex, but they can be simplified further with some assumptions. For example, the constant speed assumption, $\dot \Omega = 0 $ has the following solution
$$ \begin{aligned}
F_A & = (m \Omega^2) \frac{L^3 R_C-L^2 R_C c +L R_A c^2 + R^2 R_A c}{L(L^2-L c+R_A R_C)} \\
F_B & = (m \Omega^2) \frac{R (L-c)(L\,c+R^2)}{L(L^2-L c+R_A R_C)} \\
P_A & = (m \Omega^2) \frac{R\,R_C (c-L)}{L^2-L c+R_A R_C} \\
\end{aligned}$$
To see if this is even remotely correct, I created a small angle approximation for the general case (small steering angle $\delta$) to find that
$$ \begin{aligned}
F_A & \approx m \frac{c v_B^2-(L^2-c^2) a_B}{L^2} \sin \delta \\
F_B & \approx m \frac{(L-c) v_B^2 + (L^2-c^2) a_B}{L^2} \sin \delta \\
P_A & \approx m a_B \cos \delta \\
\end{aligned}$$
which kind of makes intuitive sense, in terms of how speed $v_B$ and acceleration $a_B$ might affect the tire forces.
and for the non-accelerating case
$$ \begin{aligned}
F_A & \approx m \frac{c\, v_B^2}{L^2} \sin \delta \\
F_B & \approx m \frac{(L-c) v_B^2 }{L^2} \sin \delta \\
P_A & \approx 0 \\
\end{aligned}$$
which is a result that matches the general rule of the best handling cars are the ones with $c = L/2$. Use this above to find $F_A \approx F_B$, which means the tires are equally loaded resulting in the best handling limit (This is the topic of different discussion).
