Lie derivative and vector field notion.

[01030312]

Here is an approach for lie derivative. And i would like to know how wrong is it.
Assuming lie derivative of a vector field measures change of a vector field along a vector field, take a coordinate system, xⁱ , and the vector field fⁱ along which Tⁱ is being changed. I go this way, i take the vector 'T' at point xⁱ, transform the coordinate as- xⁱ -> xⁱ + efⁱ= x'ⁱ, e nearly zero. Now Tⁱ(x) is transformed to T'ⁱ(x').
Now I subtract T'(x') with T(x).
We would do the same thing to find directional derivative of a scalar function- S( x + ef)- S(x)= S(x')-S(x)
Now back to vector T. T'^j(x')= Tⁱ(x)dx'^j/dxⁱ.
Now take Tⁱ(x') = Tⁱ(x) + ef^k$$\nabla$$_kTⁱ. Since T(x) is a vector field. But I don't ask for a definition of connection. We use arbitrary connection and try to see if T(x') is completely specified by T(x) by requiring specification of connection $$\nabla$$.
So T'^j(x')= Tⁱ(x') - e( f^k$$\nabla$$_kTⁱ(x) - T^k(x')dfⁱ /dx^k
To first approximation.
This finally gives, considering first order in e,
T'^j(x')= Tⁱ(x) - e( f^k$$\nabla$$_kTⁱ(x) - T^k(x')$$\nabla$$_kfⁱ)
= Tⁱ(x) + e(lie derivative required) + e²(terms..)
As is seen here, the final form does not include the notion of a definite vector field for T. Even connection $$\nabla$$ can be removed from lie derivative for vector. I need to know if it is correct, the assumption that lie derivative can be considered as change in tensor, not necessarily a tensor field. Or is a precisely defined tensor field property already used here?

 

[01030312]

I really need to know if its correct. Its simple maths though looks long. Could someone help? Experts' comments?