The overwhelming preference to work with $k$-covector fields ("differential forms") stems from a few basic facts:
First, you might know of $\nabla$ from vector calculus. It is related to the exterior derivative $d$ in the sense that you can do $\nabla \wedge$ on a covector field and it is equivalent to $d$. $\nabla$ itself transforms as a covector does, and so it takes 1-covectors to 2-covectors, $k$-covectors to $k+1$-covectors (and these are all fields, of course). So there is a very convenient element of closure under the operation.
Second, integration on a manifold naturally involves the tangent $k$-vector of the manifold. This is something traditional differential forms notation tends to gloss over. When you see, for example, something like this:
$$\int f \, \mathrm dx^1 \wedge \mathrm dx^2$$
It really means this:
$$\int f \, (\mathrm dx^1 \wedge \mathrm dx^2)(e_1 \wedge e_2) \, dx^1 \, dx^2$$
For this reason, the basis covectors $\mathrm dx^i$ should not be confused with the differentials $dx^i$. Further, that we use $e_1 \wedge e_2$ here, and not $e_2 \wedge e_1$, reflects an implicit choice of orientation, which is usually picked by convention from the ordering of the basis, but this need not always be the case. The tangent $k$-vector, and especially its orientation, must necessarily be considered in these integrals.
So why does this make $k$-covector-fields preferred? Because the action of these fields on the manifolds' tangent $k$-vectors is inherently nonmetrical. So, differential forms allows you to do a lot of calculus without imposing a metric.
This point, however, is somewhat obfuscated when you introduce the Hodge star operator and interior differentials, for these are metrical. Then, you get a big problem with differential forms: by working exclusively with $k$-covector fields, and expunging all reference to $k$-vector fields, the treatment when we do have a metric is extremely ham-fisted. Yes, you can do everything with wedges, exterior derivatives, and Hodge stars. But it makes much more sense to use corresponding grade-lowering operations and derivatives instead. Geometric calculus does this, but let met get to that in a moment.
Regarding the pushforward vs. pullback, I must confess a lack of understanding. I do not see why we would want to pull covectors back from a target manifold while insisting we must push vectors forward. I'm very familiar with the mathematics: that under a smooth map, the adjoint Jacobian transforms covectors from the target cotangent space to the original, and the inverse Jacobian does the same for vectors. Perhaps it has to do with defining the pushforward as the inverse of this inverse.
Now, do all these remarks put together mean that $k$-vector fields are inherently disadvantaged, or less rich, than $k$-covector fields? I would say no. I mentioned geometric calculus earlier: it is the originator of the $\nabla \wedge$ notation that I used earlier, and it handles $k$-vector fields just fine. Geometric calculus is the calculus that goes with clifford algebra, and you may find it illuminating. Many of the theorems and results of differential forms translate to geometric calculus and to $k$-vector fields. Stokes' theorem? Used extensively. de Rham cohomology? Most of the same results apply.
My point above about differential forms integrals using tangent $k$-vectors implicitly? That comes from geometric calculus, too, where the tangent $k$-vector is not digested "trivially" and you have to look at all the metrical ways in which it might interact with the vector field you're integrating.
A grade-lowering derivative is natural to use with $k$-vector fields. In geometric calculus, this is notated as $\nabla \cdot$. You can see that successive chains of $\nabla \cdot$ continually lower the grade of a field, just as successive exterior derivatives raise it.
My ultimate point is that, when you do have a metric, it's quite nonsensical to treat everything as a differential form instead of using $k$-vector fields when appropriate. I feel the tendency to do this in physics divorces students from a lot of the vector calculus they had learned, unnecessarily so. I can't speak to mathematics courses, but I imagine some of that criticism applies, too.
Now, there are some properties of covector fields and exterior derivatives that are nicer than working with vector fields. For instance, under a map $f(x) = x'$ with adjoint Jacobian $\overline f$, it's true that $\overline f(\nabla ' \wedge A') = \nabla \wedge A$ for some covector field $A$. That's a very convenient result, and there's no correspondingly nice identity for vector fields.
Best Answer
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 $i^*(dy)=cos(\theta) d\theta$