Let $F:M\to \Bbb R^n$ be the inclusion map and $dF: TM\to T\Bbb R^n$ the smooth map induced by $F$,
then we have $$dF:TM\to T\Bbb R^n,$$
$$(x,v)\mapsto (x,v).$$
$\forall x\in M$, we choose a smooth chart containing $x$ on $M$,
then $dF$ has the following coordinate representation in terms of natural coordinates for $TM$ and $T\Bbb R^n$:
$dF(x^1,\cdots,x^m,v^1,\cdots,v^m)=(F^1(x),\cdots,F^n(x),\frac{\partial F^1}{\partial x^i}(x)v^i,\cdots,\frac{\partial F^n}{\partial x^i}(x)v^i).$
We composite $dF:TM\to T\Bbb R^n$ with $T\Bbb R^n\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$, then
we get $\Phi: TM\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$. Correspondingly, $\Phi$ has the following coordinate representation:
$\Phi(x^1,\cdots,x^m,v^1,\cdots,v^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x)v^i)^2$.
Suppose $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=1$.
Because we have
$\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^1}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$
$\qquad \qquad \qquad \vdots$
$\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^m}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$
then $v_0^1\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)+\cdots+v_0^m\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=2$,
so at least one of $\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m),\cdots,\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is not equal to $0$, then $(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is a regular point of $\Phi$ such that $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=1$, hence $\Phi^{-1}(1)$ is a regular level set.
By Corollary 5.14(Regular Level Set Theorem) of Introduction to Smooth Manifolds by Lee, $UM=\Phi^{-1}(1)$ is an embedded $(2m-1)$-dimensional submanifold of $TM$, thus an embedded $(2m-1)$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$.
The counterexample just shows that two diffeomorphic smooth structures on the same set $X$ do not need to share a common atlas. However, in any case, two diffeomorphic structures cannot be distinguished in the smooth category, in the sense that every true statement in the smooth category still remains true after replacing one structure with the other one.
This also happens in the topological category: it is possible to have two homeomorphic, but different, topological structures on a set $X$. For instance, the two topologies on $\mathbb{R}$ given by $$\mathcal{T}_1 = \{[a, \, + \infty) \; : \; a \in \mathbb{R} \cup \{-\infty\} \}, \quad \mathcal{T}_2 = \{(- \infty, \, a] \; : \; a \in \mathbb{R} \cup \{+\infty\} \}$$
are clearly different. However, the map $$f \colon (\mathbb{R}, \, \mathcal{T}_1) \to (\mathbb{R}, \, \mathcal{T}_2), \quad x \mapsto -x$$
is a homeomorphism, so from the topological point of view the two spaces have exactly the same properties.
Best Answer
First of all, the $F$ you have defined is not actually a map on the tangent bundle of the manifold.The point is,when you are explicitly using coordinates to define a map,what you need to check is whether such a thing is independent of the chart chosen.The reason is that, the map you are actually defining takes a point on the manifold and gives you an output, but the same point on the manifold may have different coordinate representations.That is the reason why you cannot just use a chart and work your way with respect to it, it simply might not give you a map on the manifold at all.In your case, for example, if I use a different local chart at that point, then a point on $TM$ would end up having two different coordinate representations, what will $F$ be then?
Edit :- I thought I need to mention this.Another very common mistake with using charts to define maps is, a 'proof' that all smooth $n$-manifolds are orientable.Indeed, one just takes the 'everywhere positive $n$-form' given by $\omega = dx^1\wedge dx^2\wedge ..\wedge dx^n$.Spot the error.
:-)