I'll answer the particular question that Cecilia ultimately asked, after she elaborated her question in response to our comments.  

I believe that she was asking this:  Given a Riemannian $n$-manifold $(M^n,g)$, when does there exist a metric $\hat g$ on $TM$ such that each point in $M$ lies in some coordinate chart $x:U\to\mathbb{R}^n$ such that, in the associated tangential coordinate chart $(x,v):TU\to\mathbb{R}^n\times\mathbb{R}^n$, one has
$$
\hat g = g + {}^tdv\circ dv.
$$
Let's say that such a coordinate chart $(U,x)$ is *adapted* to $\hat g$.

The key to answering this question is to consider what happens on overlaps.  If $(U,x)$ and $(V,y)$ are two adapated coordinate charts, then one has $y = F(x)$ for some $F:x(U)\to y(V)$ and the associated tangential coordinates are related by $(y,w) = \bigl(F(x),F'(x)v)$.  Then the requirement that, on $TU$, one have
$$
g + {}^tdv\circ dv = \hat g = g + {}^tdw\circ dw
$$
implies that ${}^tdw\circ dw = {}^tdv\circ dv$, which implies that ${}^t(F'(x))\ F'(x)$ be the identity for all $x$.  In particular, $F'(x)$ is orthogonal for all $x$ and this implies that $F'(x)$ is locally constant.  Conversely if $F'(x)$ is locally constant and orthogonal for all $x$, then one has ${}^tdw\circ dw = {}^tdv\circ dv$.

In particular, the atlas of $\hat g$-adapted coordinates $(U,x)$ consists of charts whose transition functions are locally Euclidean isometries, and it follows that $M$ must admit the structure of a flat Riemannian manifold (even though $g$ itself need not be flat).  

Conversely, if $M$ admits the structure of a flat Riemannian manifold, say $h$ is a flat Riemannian metric on $M$, then, letting $\hat h$ be the natural induced flat metric on $TM$, one can set $\hat g = \pi^\ast(g-h) + \hat h$, where $\pi:TM\to M$ is the basepoint projection, and this will construct a $\hat g$ for which the atlas of $h$-isometric coordinate charts is the atlas of $\hat g$-adapted coordinates.