Let us write the isomorphism
$T_{(x,v)}TM = H_{(x,v)}TM \oplus V_{(x,v)})TM \cong T_xM \oplus T_xM$
by
$\xi \simeq (\xi^h,\xi^v)$,
so that $\xi^h \in T_xM$ and $\xi^v \in T_xM$.
Here the identification $H_{(x,v)}TM \cong T_xM$ is given by the restriction of $d_{(x,v)}\pi$ to $H_{(x,v)}TM$ (where $\pi:TM \rightarrow M$ is the projection), and the isomorphism $V_{(x,v)}TM \cong T_xM$ is canonical.
The key point is that under these identifications, if $z$ is a curve on $TM$, say $z(t)=(\gamma(t),u(t))$ then
$\dot{z}(0) \simeq (\dot{\gamma}(0),(\nabla_tu)(0))$.
So suppose $x \in M$ and $v,w,y \in T_xM$. Let $\gamma$ be a curve in $M$ such that $\gamma(0)=x$ and $\dot{\gamma}(0)=w$, and let $u$ be a vector field along $\gamma$ such that $u(0)=v$ and $(\nabla_tu)(0)=y$. Let $z(t)=(\gamma(t),u(t))$.
Think of $A$ as a map $TM \rightarrow TM$, so that the differential $dA$ is a map
$d_{(x,v)}A:T_{(x,v)}TM \rightarrow T_{(x,v)}TM$.
Then given $w\in T_xM$, if $\xi_w$ is the unique vector whose horizontal component is $w$ and whose vertical component is zero (i.e. $\xi^h = w$ and $\xi^v = 0$), then we define
$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi_w))^v$,
and similarly if $\zeta_w$ is the unique vector whose horizontal component is zero and whose vertical component is $w$ (i.e. $\zeta^h = 0$ and $\zeta^v = w$), then we define
$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\zeta_w))^v$.
Then it follows that
$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))$,
and these two maps have the properties you're looking for.
The answer that you want, namely div $U_g$, is not going to expressible in terms of a geometrically invariant quantity (such as, say, the scalar curvature of $g$) because it depends on the underlying coordinates $x$ in which you have presented the metric. For example, if $\det g(x)$ is constant (which is a coordinate dependent thing), then the divergence of $U_g$ (computed in the $xu$-coordinates, which, I assume, is what you mean by the 'Euclidean divergence') will vanish identically (and conversely, as a matter of fact).
I'll try to explain this in the symplectic formulation, since that's the version I find the clearest, but I think that you can make the translation on your own. You start with a Riemannian metric $g = dx\cdot G(x) dx$, where $G$ is a function on $\mathbb{R}^n$ with values in positive definite symmetric matrices. Let's write $G = F^TF$, where $F$ is invertible (but not necessarily symmetric; you can take $F$ to be the positive definite square root of $G$ if you like, but that's not necessary for my argument). The Lagrangian is then $L = \tfrac12 u\cdot u$, where $u = F(x)\ dx$, regarded as an $\mathbb{R}^n$-valued function on the tangent bundle of $\mathbb{R}^n$ (i.e., an orthonormal coframing of the underlying manifold). Applying the Legendre tranform, the symplectic form on the cotangent bundle pulls back to the tangent bundle to be
$$
\Omega = dp\wedge dx = d((F(x)\ u)^T) \wedge dx
= (du)^T\ F(x)^T \wedge dx + u^T\ d(F(x)^T)\wedge dx
$$
The vector field $U_g$ is the $\Omega$-Hamiltonian vector field associated to $L$, i.e., it satisfies
$$
\iota(U_g)\ \Omega = -dL = - u\cdot du
$$
(where $\iota(X)$ denotes interior product, what I normally call 'lefthook'). By the standard identity, the divergence of $U_g$ with respect to the Liouville volume form, i.e., $\mu = \tfrac1{n!}\Omega^n$, vanishes identically. Now, you can compare the Liouville volume form with the 'Euclidean' volume form in $xu$-coordinates by noting that, by exterior algebra, we have
$$
\tfrac1{n!}\Omega^n = \det(F(x))\ \tfrac1{n!}\bigl((du)^T\wedge dx\bigr)^n
$$
It follows that the 'Euclidean' divergence of $U_g$ in the $xu$-coordinates is given by the formula
$$
-U_g\bigl(\log|\det(F(x))|\bigr) = -\tfrac12\ U_g\bigl(\log|\det(G(x))|\bigr).
$$
Since, as you have already computed, $U_g(x) = F^{-1} u$, it follows that the divergence you want is, as a function, linear in the $u$-coordinates. You might write it as
$$
-\tfrac12 \nabla \bigl(\log|\det(G(x))|\bigr)\cdot (F^{-1} u).
$$
Best Answer
There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.
First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $v$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$, and let $\nu_v : T_v(TM) \to T_v(TM)$ be the nilpotent endomorphism $\nu_v = \iota_v\circ \pi'(v)$.
Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl(\nu_v(w)\bigr) $$ for all $w\in T_v(TM)$. Let $R$ be the vector field on $TM$ that is tangent to the fibers of $\pi$ and that is the natural radial vector field on each such (vector space) fiber, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous on each $\pi$-fiber, then, by Euler's relation, $E_L = L$.)
Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$, consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. Then $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.
This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.