,Although messy, we can still do it by purely chart representation.
Let $w=\Sigma a_I dx^I=\Sigma_J b_J dy^J$ and assume $w$ is $k$-form and the manifold has dim $n$. Then since they are equal , we can find:
$b_J=\Sigma_I \det (\partial x^I/\partial y^J) a_I $, where $\partial x^I/\partial y^J$ denotes the jacobian matrix of $x^{i_1},...,x^{i_k}$ for $i_1<...<i_k$ with respect to $y^{j_1},...,y^{j_k}$ where $j_1<...<j_k$ .
Then we take the exterior derivative definition, we want to prove the following:
$$\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}}d{x^i} \wedge d{x^I}} } = \sum\limits_J {\sum\limits_j {\frac{{\partial {b_J}}}{{\partial {y^j}}}d{y^i} \wedge d{y^J}} } $$.
Substitute the above relationship into the right hand side , we obtain :
\begin{array}{l}
\sum\limits_J {\sum\limits_j {\frac{{\partial {b_J}}}{{\partial {y^j}}}d{y^i} \wedge d{y^J}} } = \sum\limits_J {\sum\limits_j {\sum\limits_I {((\frac{{\partial {a_I}}}{{\partial {y^j}}})} } } \det (\frac{{\partial {x^I}}}{{\partial {y^J}}}) + {a_I}\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}})d{y^j} \wedge d{y^J}\\
= \sum\limits_J {\sum\limits_j {\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}} \cdot \frac{{\partial {x^i}}}{{\partial {y^j}}}\det (\frac{{\partial {x^I}}}{{\partial {y^J}}})d{y^j} \wedge d{y^J}} } } } + \sum\limits_J {\sum\limits_j {\sum\limits_I {{a_I}\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } }
\end{array}
By the chain rule and also interchanging the order of summation, we see the first term is just equal to $\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}}d{x^i} \wedge d{x^I}} } $ and we only need to show the second term is zero.
Due to $a_I$ arbitrary, we shall prove for any fixed $I$ $$\sum\limits_J {\sum\limits_j {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {x^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } = 0.$$
Since if $j$ lies in the ordered $J$, $dy^j\wedge dy^J$ is zero, so we can do it for every $U$ subset of $\{1,2,...,n\}$ where $card(U)=k+1$.
$$\sum\limits_J {\sum\limits_j {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } = \sum\limits_{U \subseteq {\rm{ \{ 1,2,}}...{\rm{,n\} ,}}U{\rm{ has }}k + 1{\rm{ elements}}} {\sum\limits_{j \in U} {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } $$ where for every $U$ selected and for each $j\in U$ selected $J$ is uniquely determined by requiring $J\subset U$. And we can show for each summand $U$, it is zero.
Let us denote $I=(i_1,...,i_k)$ with $i_1<...<i_k$ and $U$=$\{j_1,...,j_k,j_{k+1}\}$ where $j_1<...<j_k<j_{k+1}$, for each pick of $j$, $J$ is uniquely determined. Then we can see by definition of determinant and the differentiation formula of determinant , the second order derivative term related to $i_1$ is given as the determinant of the following matrix when $j=j_1$ is picked: \begin{array}{*{20}{c}}
{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_2}}}\partial {y^{{j_1}}}}}}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_3}}}\partial {y^{{j_1}}}}}}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_4}}}\partial {y^{{j_1}}}}}}&{...}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_{k + 1}}}}\partial {y^{{j_1}}}}}}\\
{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_4}}}}}}&{...}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_{k + 1}}}}}}}\\
{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_4}}}}}}&{...}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_{{\rm{k + 1}}}}}}}}}\\
\vdots & \vdots & \vdots & \vdots & \vdots \\
{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_4}}}}}}& \cdots &{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_{k + 1}}}}}}}
\end{array} and we collect the terms for other value of $j$ similarly and note that by $dy^{j}\wedge dy^{J}$ it changes sign once when $j$ are incremented as $j_1,...,j_{k+1}$ in order.
Now let us observe the result by expanding of determinant along the first row using Laplace formula: we can easily find that the term related to $\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}}$ can be represented as $${( - 1)^{p + 1}}\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}} \cdot \det (\frac{{\partial ({x^{{i_2}}},{x^{{i_3}}},...,{x^{{i_k}}})}}{{\partial ({y^{{z_1}}},{y^{{z_2}}},...,{y^{{z_{k - 1}}}})}}) \cdot {( - 1)^r}$$ where the sign term $(-1)^{p+1}$ is due to $dy^{j}\wedge dy^{J}$ and the second sign change is due to the use of Laplace formula of determinant and it is not hard to find that we can have $r=q+1$ when $q<p$ and $r=q$ when $q>p$ and also $z_1,...,z_{k-1}$ are just indices in $\{j_1,...,j_{k+1}\}$ excluding $j_p,j_q$ and of course $z_1<....<z_{k-1}$ is required.
Now we can see that if we set $j=j_p$ and we can pick a $j_q$ in $J$ and then we can also set $j=j_q$ and pick $j_p$ for $J$ accordingly in the above summation for $U$ and they have different sign but since $\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}} = \frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_q}}}\partial {y^{{j_p}}}}}$, thus the summation is zero and thus there is no term related to the second derivative of $x^{i_1}$ and similar arguments can be applied to $x^{i_2},...,x^{i_k}$(easily by just interchanging the rows of the jacobian matrix).
Thus we conclude the proof.