Here is an answer based on the following fact, known (at least in the French folklore) as the Hadamard-Lévy theorem: if a $C^1$ map $X:\mathbb R^n\to \mathbb R^n$ satisfies $\operatorname{det}(D X(y))\neq 0$ (local invertibility) and is proper (inverse images of compact sets are compact, equivalently $|X(y)|\to +\infty$ as $|y|\to+\infty$) then it is a global diffeomorphism. So the whole argument amounts to check that indeed $y\mapsto X(s,y)$ is proper for all $s<\bar T$.

To this end, two observations are crucial:

1. the matrix $M(s,y):=I+s D^2H(Dg(y))D^2g(y)$ is symmetric
2. for all $s<\bar T$ its smallest eigenvalue is bounded away from zero uniformly in $y$, hence $M(s,y)\geq \lambda(s) I >0$ for all $y\in \mathbb R^n$ in the sense of symmetric matrices, $\lambda(s)$ being the smallest eigenvalue.

1. is clear because Hessian matrices are symmetric, and 2) follows almost immediately from the boundedness of the two Hessians as in your hypotheses (This is not difficult to check so let me omit the details here). As a consequence $X(s,.)$ grows at least linearly at infinity (with growth rate $\lambda(s)\to 0$ as $s\nearrow \bar T$), it is therefore a proper map and thus a global diffeomorphism.

Here is an answer based on the following fact, known (at least in the French folklore) as the Hadamard-Lévy theorem: if a $C^1$ map $X:\mathbb R^n\to \mathbb R^n$ satisfies $\operatorname{det}(D X(y))\neq 0$ (local invertibility) and is proper (inverse images of compact sets are compact, equivalently $|X(y)|\to +\infty$ as $|y|\to+\infty$) then it is a global diffeomorphism. So the whole argument amounts to check that indeed $y\mapsto X(s,y)$ is proper for all $s<\bar T$.

To this end, two observations are crucial:

1. the matrix $M(s,y):=I+s D^2H(Dg(y))D^2g(y)$ is symmetric
2. for all $s<\bar T$ its smallest eigenvalue is bounded away from zero uniformly in $y$, hence $M(s,y)\geq \lambda(s) I >0$ for all $y\in \mathbb R^n$ in the sense of symmetric matrices, $\lambda(s)$ being the smallest eigenvalue.

1. is clear because Hessian matrices are symmetric, and 2) follows almost immediately from the boundedness of the two Hessians as in your hypotheses (This is not difficult to check so let me omit the details here). As a consequence $X(s,.)$ grows at least linearly at infinity (with growth rate $\lambda(s)\to 0$ as $s\nearrow \bar T$), it is therefore a proper map and thus a global diffeomorphism.

Edit 1: after looking up English versions of the Hadamard-Lévy theorem, it seems that the latter is known more generally as the Hadamard's global inverse function theorem or sometimes Hadamard-Cacciopoli theorem, see this post on Math.SE (and in particular the first answer).

Edit 2: oooops, the product of two symmetric matrices is NOT symmetric in general, so the argument has a "tiny" gap. I'll leave my answer as is, since I suspect that the issue can be fixed relatively easily and that the line of thought should be correct somehow.