A kinetically-favored pathway will have the lower activation barrier. A reaction that has a lower barrier will proceed faster, as per the classical Arrhenius equation. However, a kinetically-favored pathway will not produce products with the lowest free energy in the system. That's what makes it a kinetic barrier.
So how do we understand this? Think about kinetic versus thermodynamic enolates. Let's use 3-methyl 2-butanone as an example. If you want to make a kinetic enolate, you cool the reaction down and use a sterically-bulky base like t-butoxide. It will attack the least sterically-hindered alpha carbon to generate the enolate. It faces a low activation barrier because the base and substrate doesn't have to orient themselves in a very specific way - those protons are just sticking out for the world to see. So since the transition state doesn't need to be very ordered, it's lower in energy. Compare that to the thermodynamically-favored pathway, where the base has to wedge itself into a very sterically-crowded environment. This is energetically costly and the transition state therefore is raised in energy relative to that of the kinetic pathway. This has the effect of decreasing rate.
