The previous answer to this question is actually incorrect, the last contraction with $V_b$ doesn't make sense mathematically. The formulation of the problem misses one crucial detail present in <PERSON> p. 72, the killing vector field $K$ is orthogonal to a family of spacelike surfaces. This introduces an additional condition coming from <PERSON>' theorem $$ K_{[a;b} K_{c]} = 0. $$ This identity gives the missing piece $K_{[a;b]}$, which is not constrained by the killing equation $$ K_{a;b} + K_{b;a} = 0. $$ Using both the <PERSON>' theorem and the Killing equation we indeed can reproduce the result from <PERSON>. Indeed, using the Killing property we get $$ K_{a;b} K^b = - K_{b;a} K^b = - \frac{1}{2} (K_b K^b){,a} = \frac{1}{2} (f^2){,a} = f f_{,a}. 768
$$ From that we get another identity $$ K^a K^b K_{a;b} = f f_{,a} K^{a} = K_{(a;b)} K^a K^b = 0, $$ because, again, $K$ is a Killing vector. From that we see that $$ f_{,a} K^a = f_{,a} V^a = 0. $$ We are ready to establish the identity for $\dot{V}a$ $$ \dot{V}_a = V{a;b} V^b = (f^{-1} K_{a}){;b} V^b = f^{-2} K{a;b} K^b - f^{-2} K_a f_{,b} V^b = f^{-1} f_{,a}. $$ Finally we take the Frobenius' theorem, using the Killing equation we can rewrite it in the following way $$ K_{a;b} K_c + K_{b;c} K_a + K_{c;a} K_b = 0. $$ We contract this identity with $K^c$ to get $$ K_{a;b} f^2 = K_{b;c} K^c K_a + K^c K_{c;a} K_b = f f_{,b} K_a - f f_{,a} K_b = f^3 (\dot{V}b V_a - \dot{V}_a V_b). $$ Finally, $$ V{a;b} = (f^{-1} K_a){;b} = f^{-1} K{a;b} - f^{-2} f_{,b} K_a = \dot{V}_b V_a - \dot{V}_a V_b - \dot{V}_b V_a = -\dot{V}_a V_b. $$
I spent quite a lot of time trying to figure out the missing piece, <PERSON>' theorem, it is surprisingly easy to miss. I hope I'll save someone else that time. 804