It is possible to see the cotangent bundle of the submanifold as a kind of symplectic reduction of the cotangent bundle of the ambient manifold. I think it might be enough to explain the analogous fact from linear algebra.
Let V be a vector space and U a subspace. There is a natural symplectic form $\omega_V$ on $V^*\oplus V$ given by
$$
\omega_V((\alpha,u),(\beta,v)) = \alpha(v) - \beta(u)
$$
where greek letters are elements of $V^*$ and roman letters are elements of $V$. (This is just d of the Louville form in this situation.) There is an analogous form $\omega_U$ on $U^* \oplus U$.
Now, let $U^0$ denote the annahilator of $U$ in $V^*$. Consider the subspace
$$
U^0 \times\{0\} \subset V^* \oplus V
$$
This subspace is isotropic for $\omega_V$. Its symplectic complement is the coisotropic subspace $V^*\oplus U$.
Now it is a standard fact in symplectic geometry that if you divide a coisotropic subspace by its symplectic complement the result is naturally a symplectic vector space. (This is the linear algebra behind symplectic reduction.) Applying this idea here we see that the quotient
$$
(V^*\oplus U )/ (U^0\times\{0\})
$$
inherits a natural symplectic structure. Of course, the quotient is precisely $U^*\oplus U$ and the symplectic form is nothing but $\omega_U$.
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.
Best Answer
You may find an elegant proof of this fact on Paternain's book "Geodesic Flows" (Birkhauser), in the very first pages. For convenience, I will reproduce the main parts of the argument here:
The most important step is to understand the geometry of $TTM$, the tangent bundle to the tangent bundle of $M$. Henceforth denote $\pi:TM\to M$ the footpoint projection. Note that, along the zero section, there is a canonical identification of $T_{(x,0_x)}TM=T_xM\oplus T_xM$, nevertheless this is not the case for arbitrary points in $TM$. Existence of a canonical horizontal complement to the vertical space $\ker d\pi$ is equivalent to having a connection.
Connection map:
Fix a connection and consider the connection map $K:TTM\to TM$ defined as follows. Consider $\xi\in T_\theta TM$ and a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$. These give rise to a curve $\alpha=\pi\circ z$ (the projection of $z$ onto $M$)and a vector field $Z$ along $\alpha$, s.t. $z(t)=(\alpha(t),Z(t))$. Then $K$ is defined by $$K_\theta (\xi)=(\nabla_{\dot\alpha} Z)(0).$$
Horizontal lift:
Now, define the horizontal lift $L_\theta:T_xM\to T_\theta TM$ as follows: given $v\in T_xM$ and $\beta:(-\epsilon,\epsilon)\to M$ a curve s.t. $\beta(0)=x$ and $\dot\beta(0)=v$, let $W(t)$ be the parallel transport of $v$ along $\beta$ and $\sigma:(-\epsilon,\epsilon)\to TM$ be the curve $\sigma(t)=(\beta(t),W(t))$. Then set $$L_\theta(v)=\dot\sigma(0)\in T_\theta TM.$$
Finally, we recall that in the above language, the geodesic vector field $G:TM\to TTM$ is clearly given by $G(\theta)=L_\theta(v)$.
Symplectic structure of $TM$:
One can verify that, in the above notation, the canonical symplectic structure of $TM$ can be invariantly written as $$\omega_\theta(\xi,\eta)=g(d_\theta \pi (\xi),K_\theta(\eta))-g(K_\theta(\xi),d_\theta\pi(\eta)).$$
Pf. With a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$, we have: $$d_\theta H(\xi)=\frac{d}{dt}H(z(t))\big|_{t=0}$$
$$=\tfrac12\frac{d}{dt}g_{\alpha(t)}(Z(t),Z(t))\big|_{t=0}$$
$$=g(K_\theta(\xi),v)$$
$$=g(d_\theta \pi(L_\theta (v)),K_\theta(\xi))$$
$$=g(d_\theta \pi(G(\theta)),K_\theta(\xi))$$
$$=\omega_\theta (G(\theta),\xi)\quad\square$$