Let me try:

Think of each coil of the spring as a twisted torsion bar. Assuming you are within the design operating range of the spring, the force exerted by that coil is proportional to the displacement of that coil (Hooke's law). So if my spring is 10 inches long and has 10 coils, I can think of it like 10 small springs stacked up.

Lets say I depress the 10-inch spring by 1 inch and feel 100 pounds of force, that means each of the coils was displaced by 1/10-inch and exerted 10 ponds of force. If I depress the spring 2 inches, I will feel 200 pounds, etc.; until I exceed the operating range of the spring.

Now, if we remove one coil we have a 9-inch, 9 coil spring. If I depress this spring by 1 inch, each of the coils will be displaced by 1/9-inch and the resulting force will be 11.1 pounds. However, I only have 9 coils contributing to the total force so, 11.1 x 9 = 100 pounds.

The result is that the overall spring coefficient is the same for the shorter spring, but it has less usable travel. In both cases a total displacement of 1 inch results in a force of 100 pounds. Shortening the spring does not make it stiffer, just shorter.