My reading of the question is this: we're given $H\in C^\infty(M)$ with $M$ symplectic, and we want to know whether there's a submanifold $L\subset M$, a Riemannian metric $g$ on $L$, and a symplectomorphism $T^\ast L \cong M$ under which $H$ pulls back to the norm-square function. And we want to know if $(L,g)$ is unique.
Uniqueness is easy: we recover $L$ as $H^{-1}(0)$, and $g$ as the Hessian form of $H$ on the vertical tangent bundle (determined by the symplectomorphism) along $L$.
Basic necessary conditions:
(1) $L:=H^{-1}(0)$ is a Lagrangian submanifold of $M$.
(2) $L$ is a non-degenerate critical manifold of $H$ of normal Morse index 0.
These conditions imply that a neighbourhood of $L$ embeds symplectically in $T^\ast L$, and also (by the Morse-Bott lemma) that $H$ is quadratic in suitable coordinates near $L$. These two sets of coordinates needn't be compatible, so let's replace (2) by something much stronger (but still intrinsic):
(3) There's a complete, conformally symplectic vector field $X$ (i.e., $\mathcal{L}_X\omega=\omega$), whose zero-set is exactly $L$, along which $H$ increases quadratically (i.e., $dH(X)=2H$).
I claim that (1) and (3) are sufficient. With these data, you can locate a point $x\in M$ in $T^\ast L$. Flow $X$ backwards in time starting at $x$ to obtain the projection to $L$; pay attention to the direction of approach to $L$ to get a tangent ray, and use the metric (i.e., the Hessian of $H$ on the fibres of projection to $L$) to convert it to a cotangent ray. Pick out a cotangent vector in this ray by examining $H(x)$. If I'm not mistaken, this will single out a symplectomorphism with the desired properties.
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.
Best Answer
As you noticed, you need a connection $D$ on $TM$ in order to define the splitting in $T(TM)$ and the unique vector field $\xi$ with the property that $\xi$ is horizontal and satisfies $\pi_*(\xi_X)=X$ for all $X\in TM$. Then the flow $\phi_t$ of $\xi$ is the geodesic flow of the connection $D$.
To see this, recall that from the definition of the covariant derivative, a vector field $Y$ along a curve $c$ in $M$ satisfies $D_{\dot c}Y=0$ if and only if $Y$, wieved as a curve in $TM$, is horizontal (which means that $\dot Y$ is horizontal at each point wrt the above splitting).
Now, take any $X\in TM$ and denote by $\gamma_t:=\phi_t(X)$ and $c_t:=\pi(\gamma_t)$. Since $\dot\gamma=\xi$, the curve $\gamma$ is the horizontal. Moreover, $\dot c$, wieved as a curve in $TM$, is just $\gamma$: $$\dot c_t=\pi_*(\dot\gamma_t)=\pi_*(\xi_{\gamma_t})=\gamma_t.$$ By the above, $\dot c$ is $D$-parallel along $c$.