Is it always possible to create a intuition for abstract algebra theorems?

I think the best way to understand abstractions intuitively is to study lots of examples. To understand group actions, write some down. Then think about whether each is transitive, whether it's primitive, look at the stabilizers of elements. Verify that the theorem is true; try to understand why in each particular case. Look for examples where some hypothesis fails and see whether (and why) the conclusion fails.

That learning strategy reproduces (in part) the explorations that led to the useful abstractions of group theory. Mathematicians studying the symmetries of various geometrical objects realized that they could reason about the symmetry of just about anything by inventing an abstract language whose definitions captured the essence of the properties of symmetries.

Unfortunately, often the abstractions - the definitions and theorems - take center stage in teaching and learning. Students find it hard to understand what's going on without the examples. I would tell my students that for me the category of groups consisted of those groups with which I was personally familiar. I assigned homework that called for working out examples at least as much as proving theorems.

In your own discipline (computer science) there are analogous historical trends leading from examples to abstractions: the concepts of object oriented or functional programming languages, the development of abstract tools to reason about databases.

One way to build intuition is to reframe results in terms of objects and relationships that you know of and are familiar with. Since it seems as though you're looking at group action in the context of graph theory, this might be a good route to go.

For example, consider the first theorem you state about the primitivity of a transitive group action. For a graph $X$, think of $\Omega = V(X)$ and $G=\operatorname{Aut}(X)$. Then a system of imprimitivity can be seen as a partition of $X$ into subgraphs that 'look the same.' Now, the theorem states

Assume that $\operatorname{Aut}(X)$ acts transitivity on $V(X)$, and consider $v\in V(X)$. Then $\operatorname{Aut}(X)$ has a system of imprimitivity if and only if there is some nontrivial subgroup $H$ containing $\operatorname{stab}(v)$.

This might start to seem more concrete. Since $\operatorname{Aut}(X)$ acts transitively on $X$, the subgroup that sends these subgraphs to themselves not only contains $\operatorname{stab}(v)$, but strictly contains it, since $v$ can also be mapped to any other vertex in its partition.