$TS^1$ is Diffeomorphic to $S^1\times \mathbf R$. I know this is a very basic question. But I am unable to get every detail right.

I need to show that $TS^1$ is diffeomorphic to $S^1\times \mathbf R$.

(I am using the concept of derivations to define the tangent spaces.)
I asked this of my friends and this is what, in essence, I have learned from them:
Define $f:TS^1\to S^1\times \mathbf R$ as 
$$f
\left(
p;\ \lambda\left(p_2
\left.\frac{\partial}{\partial x_1}\right|_{p}-p_1\left.\frac{\partial}{\partial x_2}\right|_p
\right)
\right)
=
(p,\ \lambda)
$$
for all $p=(p_1,p_2)\in S^1$ and all $\lambda\in \mathbf R$, and show that this is a diffeomorphism.
My problem with this is that: 
$\left(p_2
\left.\frac{\partial}{\partial x_1}\right|_{p}-p_1\left.\frac{\partial}{\partial x_2}\right|_p
\right)$ is not really a member of $T_pS_1$.
It lies in $di_p(T_pS^1)\subseteq T_p\mathbf R^2$, where $i:S^1\to \mathbf R^2$ is the inclusion map.
(Reason: The manifold structure of $S_1$ is governed by the fact that $i:S^1\to \mathbf R^2$ is a smooth embedding, and thus, a tangent vector $X_p\in T_p\mathbf R^2$ is in $di_p(T_pS^1)$ if an only if $X_p\xi=0$ for all $\xi\in \mathcal C^\infty(\mathbf R^2)$ with $\xi|S^1\equiv 0$.)
Keeping this in mind, I attempted the following:
Define $F:T\mathbf R^2\to \mathbf R^2\times \mathbf R^2$ as 
$$
F\left(p,\ a_1
\left.
\frac{\partial}{\partial x_1}
\right|_p+a_2\left.\frac{\partial}{\partial x_2}\right|_p\right)
=
(p,\ a_1, a_2)
$$
for all $p\in \mathbf R^2$ and $a_1, a_2\in \mathbf R$. We note that $F$ is a diffeomorphism.
Now define $G:\mathbf R^2\times \mathbf R^2\to \mathbf R^2\times \mathbf R$
as
$G(p, a_1, a_2)=(p, \sqrt{a_1^2+a_2^2})$. We note that $G$ is smooth.
Finally, since $i:S^1\to T\mathbf R^2$ is smooth (it is more than that), we have $di:TS^1\to T\mathbf R^2$ is also smooth.
Now $G\circ F\circ di: TS^1\to S^1\times \mathbf R$ is thus a smooth map, since composition of smooth maps is smooth.
But I am unable to show that this map is the required diffeomorphism.
Can somebody help?
 A: The right construction is the one you start with, the vector is indeed tangent to $S^1$. The tangent to $S^1$ is the line perpendicular to $p=(p_1,p_2)$, which is generated by $(p_2, -p_1)=p_2(1,0)-p_1(0,1)$ (although the opposite vector is more natural, by orientation reasons). Then, the partial derivatives are exactly:
$$
\tfrac{\partial}{\partial x_1}\big|_p=(1,0),\quad \tfrac{\partial}{\partial x_2}\big|_p=(0,1)
$$
and we get the tangent vector under consideration. This said, $f$ is indeed a diffeo (the inverse is clear). 
Thus, let us look at the identification of $T_pS^1\subset T_p\mathbb R^2$ in terms of derivations. To that end, we parametrize $S^1$ in the simplest standard way: $\varphi(t)=(\cos t,\sin t)$, and say $p=\varphi(\theta)$. Then the derivation $\frac{\partial}{\partial t}$ generates $T_pS^1$. It acts as follows:
$$
\tfrac{\partial f}{\partial t}\big|_p=\tfrac{d}{dt}f(\cos t,\sin t)\big|_\theta.
$$
Now $T_p\mathbb R^2$ is of course generated by the usual partial derivatives $\tfrac{\partial }{\partial x_1}\big|_p, \tfrac{\partial }{\partial x_2}\big|_p$, and we want to express $D=\tfrac{\partial }{\partial t}\big|_p$ in terms of those two derivations. That is, we look for coefficients $\alpha_1,\alpha_2$ such that
$$
D=\alpha_1\tfrac{\partial }{\partial x_1}\big|_p+\alpha_2\tfrac{\partial }{\partial x_2}\big|_p.
$$
Since  $\tfrac{\partial x_i}{\partial x_j}=0$ or $1$ according to $i\ne j$ or $i=j$, we see that $\alpha_i=D(x_i)$, and so:
$$
\begin{cases}
\alpha_1=\tfrac{\partial x_1}{\partial t}\big|_p=\tfrac{d}{dt}\cos t\big|_\theta=-\sin\theta=-p_2,\\
\alpha_2=\tfrac{\partial x_2}{\partial t}\big|_p=\tfrac{d}{dt}\sin t\big|_\theta=\cos\theta=p_1,
\end{cases}
$$
that is:
$$
\tfrac{\partial }{\partial t}\big|_p=-p_2\tfrac{\partial }{\partial x_1}\big|_p+p_1\tfrac{\partial }{\partial x_2}\big|_p.
$$
This is the opposite of the one you have (as I had commented), the different sign coming from the way the parametrization $\varphi$ turns around the origin.
Concerning the $G\circ F\circ di$ proposal, the map we have is in fact
$$
TS^1\to S^1\times\mathbb R:(x,u)\mapsto (x,\|u\|),
$$
which is not injective: $(x,\pm u)\mapsto (x,\|u\|)$. Thus one cannot get a diffeo there.
A: A homeomorphism $TS^1 \to S^1 \times \mathbb{R}$ is just a trivialization, which is equivalent to giving a nowhere-vanishing section $s\in \Gamma(TS^1)$. To promote it to a diffeomorphism, we just need $s$ to be smooth. The space $S^1$ is a Lie group, so we can take $s(g) = DL_g(v)$ for some fixed $v\in T_{\operatorname{id}} S^1$, where $L_g : S^1 \to S^1$ is left-multiplication (which is also right-multiplication in this case) by $g$. Unwinding the definitions here gives the explicit map in the first line of your post.
