If we have two blocks, first weighing $2 \text{ kg}$ with dimensions $10 \times 2$ and second weighting $4 \text{ kg}$ with dimensions $20 \times 2$, and place them on an inclined surface, would they both start sliding down at the same inclination angle?
Do object mass and geometry (size) affect friction force?
Nothing of the above mentioned thing matter, only the coefficient of friction between the two surfaces.
TL;DR Whether or not the block will start moving (sliding) depends only on the inclination angle $\theta$ and the static friction coefficient $\mu_s$ of the inclined surface. The block will move when the following condition is satisfied
$$\boxed{\theta > \arctan \mu_s}$$
Below I give detailed derivation for this condition.
The magnitude of static friction force is defined as
$$|\vec{f}_s| = \mu_s |\vec{n}|$$
where $\mu_s$ is coefficient of static friction and $\vec{n}$ is the normal force. Static friction force always acts in the direction opposite to the motion.
Setup a coordinate system for the inclined surface such that $\hat{\imath}$ axis points along the surface in the downwards direction, and $\hat{\jmath}$ axis is $90^\circ$ clockwise from the $\hat{\imath}$ axis, i.e. perpendicular to the surface pointing upwards.
With respect to this coordinate system, the normal and downward forces are
$$\vec{n} = w \cos\theta \hat{\jmath} \quad \text{and} \quad \vec{d} = w \sin\theta \hat{\imath}$$
where $w = mg$ is the object weight, and $\theta$ is the angle of the surface to horizontal. Try to evaluate the above vectors for extreme cases:
Therefore, the magnitude of the static friction force is
$$|\vec{f}_s| = \mu_s w \cos\theta$$
For block to start moving (sliding) along the surface, the downward force must overcome the static friction force
$$w \sin\theta > \mu_s w \cos\theta$$
which can also be written as
$$\theta > \arctan \mu_s$$
To conclude, whether or not the block will start moving depends only on the inclination angle $\theta$ and the static friction coefficient $\mu_s$ of the inclined surface.
Once the block starts moving (sliding), there is no longer static friction force but kinetic friction force which also acts in the direction opposite to the motion
$$|\vec{f}_k| = \mu_k |\vec{n}|$$
where $\mu_k$ is the coefficient of kinetic friction. Note that in general $\mu_k < \mu_s$.
I don't know what you mean by "size" but the angle $\theta$ where both blocks begin to slide, that is, where the maximum possible static friction force occurs, only depends on the coefficient of static friction, and the coefficient of static friction equals the tangent of $\theta$.
So if the coefficient of static friction is the same for both blocks, they will both begin to slide for the same angle $\theta$.
Hope this helps.
The normal force on each small segment of area may vary, depending on how the mass is distributed, but the friction force is always proportional to the normal force, so the total friction is proportional to the total normal.
