Why is the yield of substitution reactions decreased at higher temperatures?
In short, this is because both elimination and substitution occur parallely and are competing; amount of elimination product + substitution product = constant. If the elimination rate increases more than the increase in substitution rate, then the ratio of elimination product to substitution product will increase. This is what happens on heating, though the rate of both elimination and substitution increase with temperature.

Broadly, there are two types of reactions: those that are kinetically controlled, and those that are thermodynamically controlled. Most elimination-substitution reactions are exothermic and are kinetically controlled. That means that the proportion of products (elimination or substitution) depends only on the rate of elimination and substitution, and not on the thermodynamic favour for the reaction.

Given that the reaction proceeds to completion, the ratio of elimination product to substitution product is equal to the ratio of elimination rate to substitution rate. Suppose elimination and substitution rates are equal, then the products form in a 50:50 proportion.

The ratio of elimination rate to substitution rate is

$$
\frac{e^{\frac{-\Delta G_{e}^{*}}{RT}}}{e^{\frac{-\Delta G_{s}^{*}}{RT}}} = e^{\frac{\Delta G_{s}^{*}-\Delta G_{e}^{*}}{RT}}
$$

where $ \frac{-\Delta G_{e}^{*}}{RT}$  and  $\frac{-\Delta G_{s}^{*}}{RT}$ are the energy barriers for elimination and substitution respectively.

Since the barrier for elimination is more than the barrier for substitution, the above ratio increases with increase in temperature, and hence more elimination product is formed at higher temperatures.