Let $(M,g)$ be a riemannian manifold and $H$ a Lie group acting on $M$. Denote by $l \colon H \times M \to M$ and $l_h \colon M \to M$ the action of $H$ on $M$.
Now $H$ acts on $TM$ by derivations, so $$L \colon H \times TM \to TM, (h,(x,v_x)) \mapsto d_x l_h(v_x).$$
If $\mathfrak{h}$ denotes the Lie algebra of $H$ and $\xi \in \mathfrak{h}$, such that
$$\left.\frac{d}{dt}\right|_{t=0} L_{e^{t\xi}}(v_x) =0$$ Does that already imply, that $$\left.\frac{d}{dt}\right|_{t=0} l_{e^{t\xi}}(x) =0.$$
Let's abbreviate $\left.\frac{\text{d}}{\text{d}t}\right|_{t=0} l_{\exp(t\xi)}(x)$ by $l_\xi(x)$. If $f: N\to M$ is a smooth morphism of $H$-manifolds (in particular meaning that $f$ commutes with the actions of $H$), then for any $n\in N$ and any $h\in {\mathfrak h}$ you have $D_{f,n} (l_h(n)) = l_h(f(n))$. In particular, if $l_h(n)=0$, then also $l_h(f(n))=0$. If you apply this to $N:=TM$ and $f: TM\to M$ the projection, you get your desired implication.