The equation $\Delta G = \Delta H - T\Delta S$ predicts that as the temperature increases, the reaction becomes more favorable because the $T\Delta S$ term becomes larger and $\Delta G$ becomes more negative.
It does predict that $\Delta G$ becomes more negative (for example for the standard state). This also means that the maximal work available from the reaction becomes larger. If you equate favorability with $\Delta G$ or maximum work, you could also say the reaction becomes more favorable. It does not mean, however, that the equilibrium constant increases. This is because the equilibrium constant depends on both $\Delta G$ and the temperature. Both terms change because we are changing the temperature and $\Delta G$ is temperature-dependent:
$$ \mathrm{ln}(K) = - \frac{\Delta G^\circ}{RT}$$
It is fine to equate more negative $\Delta G^\circ$ with larger equilibrium constant when comparing two processes at the same temperature, though.
However, the equation $$\ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$ as well as Le Châtelier's principle, predict that as the temperature increases, K will decrease (and the reaction shift to the left, thereby becoming less favorable and $\Delta G$ becomes more positive).
This is all true except for the last bit about $\Delta G$ becoming more positive. $\Delta G$ at equilibrium concentrations will be zero at both temperatures, and $\Delta G^\circ$ will become more negative, as you correctly stated in your question (assuming we can neglect the temperature dependence of enthalpy and entropy of reaction). Again, the misconception is linking changes in $\Delta G$ to changes of K at two different temperatures.