The slip ratio depends on the speed for the car you would calculate based circumferential speed of the wheel in the frame of the car (angular velocity of wheel times radius), and the actual linear speed of the car.
The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. Let's see how this could happen.
To measure the slip, lets put little green splotches of die on the circumference of the tire spaced 1cm apart. From this we can tell how much the wheel has rotated. Now imagine a car that is accelerating. What happens to the tires? Well the road is providing a force on the tire. What does that do to the bottom of the tire? Well just imagine a stationary tire that can't rotate and you apply a force tangent to the tire. This will cause the tire to deform and the part of the tire you are apply the force to will get scrunched up in the direction of the force. Now if you force the tire to rotate against the force, the scrunched up part will go to where the "ground" (the thing applying the force) is.
This means that our little green splotches, instead of being 1 cm apart, they will be .8 cm apart. Suppose there are a total of 11 splotches. Then by the time tire turns enough for each splotch gets to its original position, the wheel has rotated a full revolution. On the other hand, the car has only moved 8cm, because each of the splotches is .8 cm apart and there are 11 of them (so 10 intervals). Now when we compare this 8cm that the car has actually moved while the wheel rotated once to the full circumference of the wheel, which is 10 cm, we conclude that the wheel has slipped.
Since there will always be some scrunching given a non-zero tangent force, you will always get slip for a nonzero tangent force.
Of course this scrunching goes to zero in the limit of an infinitely rigid wheel, which is the sort of wheel used in physics homework problems.
Now for high enough slip ratios, the wheel will actually slide across the pavement, but until you get to the this point, static friction is still in play, so the car is accelerating from static friction.
[...] and there's a large normal force on the system I've described, and it points forward, along the direction that would push the car forward
The normal force points vertically upwards (it comes into existence to balance out the weight). The force pulling sideways is not a normal force but static friction. You say it right later on, but messes up the words here.
But this scenario is only applying to the section of the wheel that is on the asphalt. So, as the wheel turns, different sections of the wheel have this effect. How does this produce a thrust forward for the car if the car keeps losing that segment of the tire that is producing a friction force with the asphalt?
How do you make a thrust when walking? You do that by pushing off from the ground with one foot, and then the other foot takes over. Each contributes with a portion of the forward thrust.
Now imagine a star. When a spike is touching the ground it pushes off from the ground in the same way. It creates thrust, acceleration, while it pushes. Then the next spike takes over.
Now imagine a star with more spikes. With many more spikes. So many that you can't distinguish them. So many placed so close that it looks like... a round object. A wheel. When this wheel rolls, each spike - which is now only a single point - pushes off from the ground during the infinitely short time it is touching.
It might be a very short time of thrust done by one point on the wheel, but it happens constantly, since another point takes over immediately. In total, there is constantly a thrust going on.
Is there a tangential force fighting against $Ff$ that I haven't included?
No. (In realistic circumstances there will be some energy loss in axles and gearing, in the flexing rubber and in a deforming surface - all such losses will slow down the wheel and are in one term called rolling resistance. But in ideal models these are neglected (just like air drag is), and only static friction is working horizontally.)
Because the motor is accelerating the tires radially, [...]
A clarification on the wording: The motor is accelerating the tires by applying a torque around the axle. This makes the tire turn. Because all the particles of the tire are stuck together, they hold on to each other inwards - radially. This is why the radial acceleration appears on them. Yes, a particle in circular motion must have a radial acceleration, but the motor is only indirectly causing that - the direct reason is rather the rigid wheel and the hold-on forces from inner particles.
[...] so each segment of the tire should have a tangential force on it that is going to be, at the moment I've described, opposite to friction?
And to the question: Yes, at the contact point where static friction happens there is a force oppositely. The wheel is applying a static friction on the ground. The ground replies with a static friction on the wheel in the other direction. This is Newton's 3rd law.
But there are no other forces on the particles of the wheel than this friction.
There are of course still the holding-on forces between particles as mentioned above, which pull every particle towards the centre - but this makes no difference in the motion, since at the contact point these are perpendicular to the friction force and direction of motion and have no influence.
In the same way there are of course also still weight and a normal force, but these also work at (or through) the contact point perpendicularly to the motion and have no influence.
What if there was no motor force on the wheels, and it just had some $\omega$? This scenario I've described in the latter sentence sounds like asking if there is a friction force acting on an object moving at a constant speed.
By "no motor force" you mean no motor torque. No motor torque means no forced push. The wheel touches the ground but does not try to push itself off from the ground, because it doesn't try to speed up the rotation. When there is no pushing off from the ground, there is no need for friction to appear to hold back against anything.
Conclusion: There is no static friction happening when the wheel rolls without accelerating. In an ideal case, it will roll with constant $\omega$ forever.
Best Answer
You're always going to need the radius, one way or another. If you are calculating from the angle of steering, you will also need the vehicle's wheelbase (the distance between the front and back tires). The radius of turning is going to be (approximately) the wheelbase length divided by the angle of steering in radians. Given the radius and the speed of the vehicle, the force acting on the tires is just the centripetal force $mv^2/r$, divided by the number of tires.
In practice I expect in a tight turn, the force is unevenly distributed, with the outer tires applying more force.