If we have a Killing tensor field $K$ of type $(0,d)$, the function $$I:SM\to \mathbb{R}, \ I(v)= K(v,\dots, v) \ \ \ \ \ \ (\ast )$$ is constant along geodesic flow. This is a well-known knowledge and a possible proof is as follows: Let $\gamma$ be an arc length parameterised geodesic. Then,
$$\nabla_{\dot \gamma} (K(\dot \gamma,\dots, \dot \gamma))= \nabla K(\dot \gamma, \dots , \dot \gamma). \ \ \ \ (\ast\ast) $$ In the formula above, $\nabla K $ has $d+1$ indices, so I substututed $d+1$ vectors $\dot \gamma$ inside. In deriving $(\ast\ast)$ I used that the geodesic has the property $\nabla_{\dot \gamma}\dot \gamma=0$. Above,
$\dot \gamma$ stays for the velocity vector of the geodesic.
Next, we observe that the Killing equation implies that $$\nabla K(\dot \gamma, \dots \dot \gamma) =0 \ \ \ \ (\ast\ast\ast) .$$
Combining $(\ast\ast) $ and $(\ast\ast\ast)$, we see that the function $I$ given by $(\ast)$ is constant along every geodesic, as its derivative along the velocity vector is zero.
Next, assume that the geodesic flow is transitive. This implies that every continuous function $\tilde I:SM\to \mathbb{R}$ constant along trajectories of the geodesic flow is constant. This implies that the corresponding Killing tensor is trivial.