One of my book says that:
Then consider a scalar field $F=x^3+y^3=x^2\cdot x+y^2\cdot y$ and a coordinate transform: $x=2u,y=2v$. Then $\Gamma_{ij}^{k}=0$.
I think that $V^1=x^2, V^2=y^2, Z^1=u,Z^2=v, Z_1=x, Z_2=y$.
So that $\frac{\partial F}{\partial u}=(\frac{\partial x^2}{\partial u}+0)x=4x^2=16u^2$.
However, by normal calculation, $F=8u^3+8v^3,\frac{\partial F}{\partial u}=24u^2$.
$16u^2\neq 24u^2$. Then where am I wrong.
This is a case of bad notation. You have a chart with coordinates $Z^j$, which gives a basis of the cotangent space $dZ^j$. Then the corresponding dual basis of the tangent space is denoted as $\bf Z_i$, usually these are denoted $\bf e_i$ or $\frac{\partial}{\partial Z^i}$.
In your case this means that with $Z^1=u=x/2$ and $Z^2=v=y/2$ you get $dZ^1=du=dx/2$ and $dZ^2=dv=dy/2$, you get the tangent space basis $\bf Z_1=\pmatrix{2\\0}$ and $\bf Z_2=\pmatrix{0\\2}$ relative to the canonical basis for the $(x,y)$ coordinate system.
Note that the cited fragments are concerned with the derivatives of vector valued functions. The derivative of scalar valued functions is unproblematic, as you can take differences of function values. But with functions that have values in the tangent space, you can not directly combine the values over different points, as they belong to different vector spaces. Thus you need the connection coefficients to relate the spaces in the given coordinate chart.
