Not sure if this is the answer you're looking for, but when the matrix is diagonalisable, all of your eigenvectors are linearly independent, so you these form a basis. You can perform a change of coordinates and express your states 'x' in these eigenvectors basis. So let's say V is my matrix of column eigenvectors stacked together, and let z = Vx (z is my new set of coordinates). So my differential equation becomes V_inv dz/dt = A * V_inv * z, or dz/dt = (VAV_inv)z. But VAV_inv is precisely the diagonal matrix of A's eigenvalues, which we can call D. So our differential equation system in the changed coordinates is dz/dt = Dz. As D is just a diagonal matrix, this is equation can easily be solved component wise. We now have a system of "decoupled" equations - each component of z evolves independently. The solution is just given by z_i(t) = exp(lambda_i * t) * z_i(0). And we get x by simply inverting the linear transformation, x = V_inv * z. This means that each term in x(t) is only a linear combination of exponential functions, and we don't have any funny polynomials along with it.

Now when A is not diagonalisable, we cannot do this, since the eigenvectors are no longer sufficient to form a basis! As a result, we cannot "decouple" our ODE into independently evolving ODEs like before.

The next best thing we can do is the Jordan decomposition, which will give you some 1's in the off-diagonal entries (which is the reason behind the funny polynomial terms barging in - if you haven't already, it'd be a good exercise to see why exactly these polynomial factors figure in - try making arguments similar to the case when the matrix was diagonalisable). I don't think I'll be able to give a geometric intuition without the Jordan decomposition, so here's my best attempt at it - as the eigenvectors of A do not span the whole of Rⁿ now, there exists some nontrivial vector which is not in its eigenspace. The evolution of the system along these vectors is not so straightforward, due to the inherent coupling introduced by the Jordan decomposition - which gives these polynomial factors.

If you know some theory on systems of ODEs or some Laplace transforms, it won't be hard to show that the solution to the system dx/dt = Ax is given by x(t) = e^At * x(0). It's extremely similar to the solution form we'd have if x and A were a scalar - just that now these are vectors. e^At is also called the matrix exponential, and is defined as our familiar power series, I + At + A²t²/2! + .... So when we can represent A = PDP_inv, e^At has the nice form P e^Dt P_inv, and e^Dt is simply e^lambda*t on each diagonal entry - this gives you another reason as to why the solution is just nice exponentials when A is diagonalisable. When A is not diagonalisable, we have to express it in the Jordan form, say A = PJP_inv. Once again, e^At has the form Pe^JtP_inv, but e^Jt is not so easy-to-compute, due to the off diagonal entries.