Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?

The key difference lies in the fact that the maps defined by a Markov matrix are discrete-time whereas the dynamical systems you refer to are continuous-time systems. In either case, a "location" on the system is stable or unstable depending on whether or not a perturbation off of the location grows or shrinks, which is measured differently in each system.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $e^{\lambda t}$ since (taking liberal approximations) $x(t+\Delta t) \approx x(t) + \lambda x(t) = (1 + \lambda)x(t)$; the function grows if $\lambda > 0$ and shrinks if the opposite.

For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $x_{n+1} = \lambda x_{n} > x_n$. Hence, $\lambda > 1$ indicates an unstable fixed point.

With discrete-time Markov processes, you’re examining the behavior of successively larger powers of a fixed matrix, which depends on successively larger powers of its eigenvalues. If there are any with modulus greater than one, those contributions to the matrix power will grow without bound, any with modulus less than one will tend toward zero, and any with modulus exactly one will neither grow nor shrink.

If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $A$ we’re interested in its exponential $e^{tA}$, which in turn depends on exponentials $e^{\lambda t}$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.