Why does heating transform the aquacomples of slightly acidified cobalt(II) chloride solution to the chlorido complex?
You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as:
$$\Delta G = \Delta H - T\Delta S\tag{1}$$

In this case, we would be observing $\Delta G$ for equilibrium $(2)$.

$$\ce{[Co(H2O)6]^2+ + 4 Cl- &lt;=&gt; [CoCl4]^2- + 6 H2O}\tag{2}$$

Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.

You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.

One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.