Induced map from circle-to-plane inclusion map I am reading “Introduction to Manifold” by L.W.Tu, and have trouble in understanding some facts given in Exercise 11.2 and Remark in Example 17.15 , which may be summarized as follows: 
Suppose the inclusion map from $i :S^1 \to \mathbb{R}^2$ and $i(\bar{x})=x$. The induced map $i_*: TS^1 \to T\mathbb{R}^2$ satisfies $i_*(\partial_{\bar{x}}) = \partial_x + (\partial\bar{y}/\partial\bar{x}) \partial_\bar{y}$, while for differential forms $i^*:T^*S^1\to T^* \mathbb{R}^2$ satisfies $i^* d\bar{x}=dx$. 
I can see that these facts are true when calculated explicitly, but I cannot understand how should I depict them. That is, when considering the vector fields or differential forms on $S^1$, they are one-dimensional and I have seen they are written with the parameter $\theta$ or like that. I cannot explain why the vector fields and differential forms should written with the induced Cartetian coordinates. How can I visualize such vector fields and differential forms on $S^1$?
Could someone help me?
Thank you in advance.
 A: A chart of $S^1$ can be $\phi^{-1}$ where 
$\phi: (0,2\pi)\to S^1$ 
$\phi(\theta):=(cos(\theta),\sin(\theta))$
Now you say that 
$i_*(\partial_\theta)=a\partial_x+b\partial_y$
where $(x,y)$ are the coordinates of $\mathbb{R}^2$ 
We want calculate $a,b\in \mathbb{R}$ but 
$a=i_*(\partial_\theta)(x)=\partial_\theta(x\circ (i\circ (\phi^{-1})^{-1}))=\partial_\theta(cos(\theta))=-sin(\theta)$ 
while 
$b=i_*(\partial_\theta)(y)=\partial_\theta(y\circ (\phi^{-1})^{-1}))=\partial_\theta(sin(\theta))=cos(\theta)$ 
So
$i_*(\partial_\theta)=-sin(\theta)\partial_x+cos(\theta)\partial_y$
You can observe that $i_*$ is an injective linear map because there is not $\theta$ such that $-sin(\theta)=cos(\theta)=0$ so $i$ is an immersion map. 
It is not correct to write $i^*:T^*S^1\to T^*\mathbb{R}^2$ because the differentital of a smooth map is  a contro-variant functor, so you must have 
$i^*:T^*\mathbb{R}^2\to T^*S^1$.
Now
$i^*(dx)=\alpha d\theta$
and $i^*(dy)=\beta d\theta$
for some $\alpha,\beta$, so 
$\alpha=i^*(dx)(\partial_\theta)=dx(i_*(\partial_\theta))=dx(a\partial_x+b\partial_y)=a=-sin(\theta)$ while 
$\beta= i^*(dy)(\partial_\theta)=dy(i_*(\partial_\theta))=dx(a\partial_x+b\partial_y)=b=cos(\theta)$
Then 
$i^*(dx)=-sin(\theta) d\theta$
and $i^*(dy)=cos(\theta) d\theta$
