# Problem with extending a vector field's action on another vector field

I'm following Schuller's lectures on gravity and light (this question specifically concerns this part of the video).

Let $$M$$ be a smooth manifold with a smooth vector field $$X$$, and let $$f\in C^{\infty}(M)$$ be a smooth function on $$M$$. $$X$$ is a $$C^{\infty}(M)\to C^{\infty}(M)$$ map so that $$f\mapsto Xf$$, where $$(Xf)(p)$$ is the directional derivative of $$f$$ in the direction of $$X(p)$$. This much is clear.

The lecturer indicates that the covariant derivative is introduced to extend the notion of derivatives of tensor fields w.r.t. a vector field, whereas $$X$$ only covers the notion of derivatives of scalar fields w.r.t. a vector field. So I tried to consider how $$XY^j$$ would look like, where $$Y^j$$ is a component of another smooth vector field $$Y$$. If $$X=X^i\partial_i$$, then (the subscript $$(x)$$ denotes that the quantity is defined under the chart map $$x$$): $$XY^j(p)=X^i(p)\cdot\partial_iY^j(p)=X^i_{(x)}(p)\cdot[\partial_i(Y^j\circ x^{-1})(x(p))] \\=X^i_{(x)}(p)\cdot\lim_{h\to 0}\frac{(Y^j\circ x^{-1})(x(p)+[0,\ldots,h,\ldots,0])-(Y^j\circ x^{-1})(x(p))}{h}$$ where $$[0,\ldots,h,\ldots,0]\in\mathbb{R}^n$$ has $$h$$ in its $$i$$-th slot.

Sometimes it's mentioned that the trouble with defining the derivative of a vector field is that we can't subtract vectors at different points. However, the above expression is valid since it just contains $$\mathbb{R}^n\to\mathbb{R}$$ maps and we can subtract real numbers. This is similar to the problem of finding derivatives of smooth functions - even though we can subtract values of functions at different points in the manifold, we can't subtract one point on the manifold from another. This is circumvented via defining the derivative via chart maps, which is what we've done above.

So there must be some other problem with the above definition of the derivative of a vector field component, but I can't figure it out. Can anyone help?

In any chart with coordinates $$x^i$$ you can certainly define the $$X$$ derivative of the components $$Y^i(x)$$ of the vector $$Y$$. This is just subtracting numbers as you say. The problem is that the resulting numbers $$Y^j_X(x)\equiv X^j \partial_j Y^i$$ are not components of a vector.
If $$X=X^i\partial_i$$ and $$Y=Y^j \partial_j$$ then $$XY = (X^i\partial_i Y^j) \partial_j +X^iY^j\partial^2_{ij}$$ and $$\partial^2_{ij}$$ is not a basis vector of the tangent space.