I conjecture that
$\xi \in T_{x,v} T^{1}M$ if only if $\langle K_{x,v}(\xi), v \rangle = 0$
with $K: TTM \to TM$ the connection map is given by $K_{x,v}(\xi) =\frac{D_{h}V}{dt}(0)$ where $V$ is some curve in $TM$ such that $V(0)=(x,v)$ and $V'(0)=\frac{dV}{dt}(0) =\xi$.
where $T^{1}M= UTM =$ the unit tangent bundle.
I have no idea how to prove it.
Hints or solutions will be appreciated.
$\def\g#1#2{\left\langle #1 , #2 \right\rangle}$The unit tangent bundle is the submanifold of $TM$ defined as the regular level set $T^1 M = f^{-1}(1)$ of the smooth function $f : TM \to \mathbb R : v \mapsto \g{v}{v};$ so its tangent space $T_{x,v} T^1 M$ is the kernel of $df|_{x,v}$. Given an arbitrary $\xi \in T_{x,v} TM,$ let $V : (-\epsilon , \epsilon) \to TM$ be a curve with $V'(0) = \xi$ and compute using the chain rule and metric-compatibility $$df|_{x,v} (\xi) = \frac{d}{dt}\Big|_{t=0}(f \circ V)= 2\g{\frac{DV}{dt}\Big|_{t=0}}{v} = 2\g{K_{x,v}(\xi)}{v}.$$ Thus the kernel is exactly the subspace you conjecture.