Change of Coordinate Formula for Differential Forms Let $M$ be a manifold, $x$ local coordinates on an open set $U$, $y$ local coordinates on an open set $V$. In addition, let $(x, \alpha)$ and $(y, \beta)$ be two induced bases for the common part of cotangent bundle $T^*M$. Now define a $2$-form on $T^*M$ by $\omega := d\alpha_i\wedge dx^i$. Write $\omega$ in $(y, \beta)$ system. I　know that $dx^i = \frac{\partial x^i}{\partial y^j} dy^j$. Then it is only left to write $d\alpha_i$ in $(y, \beta)$-coordinates. How do I　find this transformation formula, please? I always have difficulty with transformation between coordinates. Some intuition will be much appreciated. Thank you!
 A: A point in the tangent bundle consists of a pair $(p, v)$, where $p \in M$ and $v \in T_p^\ast M$. In $U$, $\{dx^1, \dots, dx^n\}$ constitutes a frame for the tangent bundle, which means that any covector $v \in T_p^\ast M$ can be uniquely written as $\sum_i \alpha_i dx^i(p)$ for $n$ real numbers $\alpha_i$. Therefore a point $(p,v)$ in $T^\ast U$, an open subset of $T^\ast M$, can be specified by specifying the $2n$ coordinates $(x^1, \dots, x^n, \alpha_1, \dots, \alpha_n)$. The $x_i$'s specify the point $p$, and the $\alpha_i$'s, which are coefficients on the $dx^i$'s, specify the covector $v$. (I assume this is you intended the $\alpha_i$'s to mean.) The $y^j$'s and $\beta_j$'s give coordinates on $T^\ast V$ in a similar way.
From the coordinates $(x^1, \dots, x^n, \alpha_1, \dots, \alpha_n)$ on $T^\ast U \subset T^\ast M$, we obtain a local frame for $\{dx^1, \dots, dx^n, d\alpha_1, \dots, d\alpha_n \}$ for $T^\ast(T^\ast M)$ (and similarly, another frame using the $y$/$\beta$ coordinates).
Now, on to your two-form $\omega = \sum_i d\alpha_i \wedge dx^i$. Remember, this is a two-form on the manifold $T^\ast M$, which means it is a map (or a "section") $T^\ast M \to \Lambda^2 T^\ast(T^\ast M)$. (Don't confuse it with a two-form on $M$, which is a map (or a "section") $M \to \Lambda^2 T^\ast M$.)
Your question is essentially: how do we write $d\alpha_i$ in terms of the $d\beta_j$'s?
Remember, $\alpha_i$ represents a coefficient on $dx^i$, and $\beta^j$ represents a coefficient on $dy^j$, so we will need to use the fact that, as you said, $dx^i = \sum_j \frac{\partial x^i}{\partial y^j} dy^j$. So if we have a covector $v \in T_p^\ast M$ that can be written $v = \sum_i \alpha_i dx^i(p)$, then
\begin{align*}
v &= \sum_i \alpha_i dx^i(p) \\
&= \sum_i \alpha_i \left( \sum_j \frac{\partial x^i}{\partial y^j}(p) \, dy^j(p) \right) \\
&= \sum_j \left(\sum_i \frac{\partial x^i}{\partial y^j}(p) \alpha_i \right) dy^j(p) \\
\end{align*}
This shows that
$$\beta_j = \sum_i \frac{\partial x^i}{\partial y^j} \alpha_i .$$
This is an equality of functions $T^\ast M \to \mathbb{R}$ (actually, I suppose the domain of these functions is $T^\ast(U\cap V) \subset T^\ast M$). Taking $d$ of both sides (this is the $T^\ast M$-$d$, which in this case takes a function $T^\ast M \to \mathbb{R}$ and gives a one-form on $T^\ast M$), we obtain (using the Leibniz rule)
$$d\beta_j = \sum_i d\left(\frac{\partial x^i}{\partial y^j}\right) \alpha_i + \sum_i \frac{\partial x^i}{\partial y^j} d\alpha_i.$$
It will be convenient to move the first term on the right to the left:
$$d\beta_j - \sum_i d\left(\frac{\partial x^i}{\partial y^j}\right) \alpha_i = \sum_i \frac{\partial x^i}{\partial y^j} d\alpha_i.$$
It will also be convenient to write $d\left(\frac{\partial x^i}{\partial y^j}\right)$ in terms of the $dy^k$'s: $d\left(\frac{\partial x^i}{\partial y^j}\right) = \sum_k \frac{\partial^2 x^i}{\partial y^j \partial y^k} dy^k$, so
$$d\beta_j - \sum_{i,k} \frac{\partial^2 x^i}{\partial y^j \partial y^k} \alpha_i dy^k = \sum_i \frac{\partial x^i}{\partial y^j} d\alpha_i.$$
Now, finally we can see how the two-form $\omega$ must transform:
\begin{align*}
\omega &= \sum_i d\alpha_i \wedge dx^i  \\
&= \sum_i d\alpha_i \wedge \left( \sum_j \frac{\partial x^i}{\partial y^j} dy^j \right) \\
  &= \sum_j \left( \sum_i \frac{\partial x^i}{\partial y^j} d\alpha_i \right) \wedge dy^j \\\
  &= \sum_j \left( d\beta_j - \sum_{i,k} \frac{\partial^2 x^i}{\partial y^j \partial y^k} \alpha_i dy^k \right) \wedge dy^j \\
  &= \sum_j d\beta_j \wedge dy^j - \sum_{i,j,k} \frac{\partial^2 x^i}{\partial y^j \partial y^k} \alpha_i \, dy^k \wedge dy^j \\
\end{align*}
But the last term is zero because partial derivatives commute and $dy^j \wedge dy^k = - dy^k \wedge dy^j$! Somewhat miraculously, we have shown that
$$\omega = \sum_i d\alpha_i \wedge dx^i  = \sum_j d\beta_j \wedge dy^j .$$
This means that $\omega$ is independent of the coordinate system and thus represents a globally-defined two-form on $T^\ast M$.
It turns out that $\omega$ is a very important object; it is the canonical symplectic two-form on the cotangent bundle, and it is the exterior derivative of the "tautological one-form" $\theta$, which is defined as 
$$\theta = \sum_i \alpha_i dx^i.$$
An easier way to check that $\omega$ transforms correctly would have been simply to check that $\theta$ transforms correctly, and thus is globally defined. $\omega$ is clearly equal to $d \theta$, so it must be globally defined. For more information, you may want to read Wikipedia.
