Suppose we have a smooth manifold $M$, and one point $p\in M$. Let $U$ be an open set around $p$ that admits two parametrizations, namely $x:V_1\to U$ and $y:V_2\to U$. Now in these two parametrizations we can write vector field $X$ on $M$ as
$X=\sum_{i=1}^n{a_i(p)\frac{\partial}{\partial x_i}}=\sum_{i=1}^n{b_i(p)\frac{\partial}{\partial y_i}}.$
Suppose now that we have a smooth function $f:M\to R$. Then value $Xf$ should not depend on the parametrization. So, I am trying to prove that.
Denote by $A(p)=(a_1(p),...,a_n(p))$, similarly for $B(p)$ and let $M=D(y^{-1}\circ x)(x^{-1}(p))$. Then we have $B(p)=M\cdot A(p)$. Denote by $\nabla_Xf$ and $\nabla_Yf$ gradients of $f$ with respect to two parametrizations. Then we get:
$Xf=A(p)\cdot \nabla_Xf$ for the first parametrization and
$Xf=B(p)\cdot \nabla_Yf=M\cdot A(p) \cdot \nabla_Xf\cdot M^{-1}$, for the second parametrization.
However these two are not equal. Could you please point out the mistake?