I am curious about the main ways the geometric intuition is different in studying schemes with cohomology (ie Hartshorne) vs. studying schemes without cohomology (ie Eisenbud-Harris).
It would be very useful to know how the types of geometric questions one is able to solve differ between the cohomological vs. non-cohomological approach. Thank you.
This is a pretty good question. A good suggestion is to look at Hartshorne's Chapters III, IV, V to see what sort of results are proved using cohomology. Here is one example of how you might be limited without using cohomology:
The Riemann-Roch Theorem for curves is one of the fundamental results in the theory of curves. The proofs of this that I am aware of use cohomology either implicitly or explicitly. (You can see Miranda's Algebraic Curves and Riemann Surfaces for a proof that sort of side-steps cohomology, but in much less generality.) As you will see by reading Hartshorne, the result follows rather quickly once one has developed enough results on cohomology. In particular, the proof of Riemann-Roch is quite short once one has proven Serre Duality. If I recall correctly, Eisenbud and Harris does not prove Riemann-Roch at all.