Must a good leaving group be a weak base, or a weak nucleophile?
It is a common misconception that good leaving groups should be weak nucleophiles. However, there is a simple counterexample: iodide is known to be both a good nucleophile and a good leaving group in $\mathrm{S_N2}$ reactions.

However, it does make sense for the leaving group to be defined in thermodynamic properties. Consider the reaction equation for a nucleophilic substitution following an $\mathrm{S_N2}$ mechanism:

$$\ce{Nu- + R3C-Lg -&gt; [Nu\bond{...}CR3\bond{...}Lg]^\ddagger -&gt; Nu-CR3 + Lg-}\tag{1}$$

Nucleophilicity, i.e. the ability of $\ce{Nu-}$ to attack the reaction centre is, as you noticed, a kinetic property: the nucleophile has to approach the electrophilic carbon atom from a certain direction in a certain angle with appropriately constructed orbitals. The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

However, the leaving group’s principal characteristic is to diffuse away after the reaction. If it is diffusing away, its kinetic properties no longer influence how the reaction proceeds as it is not attacking anything anymore. Rather, a favourable overall reaction enthalpy is achieved if $\ce{Lg-}$ is able to stabilise the negative charge well (or lone pair, if e.g. $\ce{Lg} = \ce{NR3+}$). This is indeed a thermodynamic property, not a kinetic one. And stabilising a negative charge directly correlates with being a weak base.