Pull-back of the metric tensor via the flow (one parameter group)

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Suppose now that $(, )$ is a Riemannian manifold. Let $\theta_t$ be a (local) one-parameter group of isometries of $(M, g)$ generating a vector field $X$. Then $$(\theta_t)^* = g$$

I wanna make sure I understand this correct.

So by the definition $$(\theta_t)^*g(X,Y)=g((\theta_t)_*X,(\theta_t)_*Y)=g(X(f\circ\theta_t),Y(f\circ\theta_t))$$

but I don't see how $g(X(f\circ\theta_t),Y(f\circ\theta_t))=g(X,Y)$.

Could someone explain what is going on here?