Here's maybe a different answer with some equations. I don't think your two statements are contradictory, because they are too (as you say) ``hand-wavey'' to be definitive. I think establishing a bit of the thermodynamics (equilibrium statistical mechanics) and non-equilibrium statistical mechanics is necessary to give proper definitions which show there is no contradiction between "[we can] state of the system slowly, making sure that while the state is changing (e.g. its P and V is changing with time), it remains in equilibrium" and "a system with internal temperature variations is not in equilibrium".
Within thermodynamics, pressure P(n,T) is a function of the number density $n=N/V$ and the temperature $T$ (e.g. ideal gas law $P=nT$). this known as an equation of state. Further, there are many known relations between thermodynamic variables and such relations can be arrived at via Maxwell Relations for a given ensemble (e.g. cannonical ensemble). In short, changing one quantity may mean another quantity is changed.
Using statistical mechanics, we can define these thermodynamic variables in terms of a phase space distribution $f(x,v,t)$ cite. They are given
$$n(x,t) = \int dv f(x,v,t),$$
$$u(x,t) = \frac{1}{n}\int dv \, v f(x,v,t),$$
$$T(x,t) = \frac{m}{3n} \int dv \, (v-u)^2 f(x,v,t),$$
$$\overset\leftrightarrow{P}(x,t) = m \int dv \, (\vec{v}-\vec{u}) \otimes (\vec{v}-\vec{u}) f(x,v,t).$$
Since every quantity explicitly depend on $f$, changes in density, $\partial_t n(x,t)$, temperature $\partial_t T(x,t)$, etc... are defined by $\partial_t f(x,v,t)$. The main take away is that.. we can talk about all of them at the same time, by speaking of $\partial_t f(x,v,t)$.
Okay now show there is now contradiction and hence answer your question... "What are the conditions for stable equilibrium in a reversible process?". The conditions are that entropy is not being generated. We know that Boltzmann entropy is
$$S = \int dx dv \, f(x,v,t) \log (f(x,v,t))$$
and so entropy generation is defined as
$$\partial_t S = \int dx dv \, \partial_t f(x,v,t) \left( \log (f(x,v,t)) + 1 \right) $$
So if we have $\partial_t f(x,v,t)$ such that $\partial_t S = 0$ then we can change the density, temperature, pressure, etc.. and keep the system in ``equilibrium''.
Having established a definition of equilibrium, I want to revisit the claim that 'a system with internal temperature variations is not in equilibrium'. This is a misleading heuristic because entropy generation can manifest changes in the temperature, but the temperature can also change without entropy being generated. To get a full picture of this, consider $\partial_t f$ in an ideal gas system, such that entropy is generated ($\partial_t S \geq 0$). Eventually entropy will maximize and the particles will have a Gaussian velocity distribution. (also discussed here). The temperature defines the width of that Gaussian (variance of particle velocity). So maybe temperature change occurs because of entropy generation that changes the variance of the velocity distribution, but this is not the only way that the velocity distribution variance can change (see Vlasov equation) and so changes in temperature need not imply that entropy is being generated.
