I'm reading 'Core Principles of Special and General Relativity' by Luscombe, specifically the introductory section on problems with defining usual notion of differentiation for tensor fields. I'll quote the relevant part:
The second way (to see whether the partial derivative of a tensor is a tensor) is to look at the definition of derivative, $$\frac{\partial T^i}{\partial x^j}=\lim_{dx^j\to 0}\frac{T^i(x+dx^j)-T^i(x)}{dx^j}$$ The numerator is not in general a vector! We're comparing (subtracting) vectors from different points, yet the transformation properties of tensors are defined at a point.
Since the equation above is a notational mess, here's my attempt to interpret it: $$\bigg(\frac{\partial T^i}{\partial x^j}\bigg)_p=\partial_j(T^i\circ x^{-1})(x(p))=\lim_{h\to 0}\frac{(T^i\circ x^{-1})(x(p)+[0,\ldots,h,\ldots,0])-(T^i\circ x^{-1})(x(p))}{h}$$ where $[0,\ldots,h,\ldots,0]\in\mathbb{R}^n$ has $h$ as its $j$-th coordinate.
Is my above interpretation correct? If so, what's the issue with defining the derivative of a vector field component in this way?
What you've defined is exactly how one would normally define the partial derivative of a tensor component $\partial_j T^i$, which, given a coordinate system, is perfectly well defined (locally). The problem is that we cannot easily interpret $\partial_j T^i$ as the components of a coordinate independent object.
There's a bit of a pedagogical divide in how this subject is approached in math and in physics. In physics, it's common to define everything in coordinates and define (coordnate independent) tensors as collections of component functions which transform appropriately under change of coordinates. In this view, $\partial_j T^i$ is not a tensor because it does not transform properly. When Luscombe writes that an obeject is "not a vector" they mean that if cannot be interpreted as the components of a tangent vector defined independently of coordinates.
In mathematics, it's more common to define things in a coordinate independent way first, and only afterword write out the coordinate representation.
As an example, take the differential of a smooth function $f\in C^\infty M$. It can be defined as a covector field whose action on tangent vectors is given by $df(v):=v(f)$, which is manifestly coordinate independent. In coordinates, we would have $$ df=\partial_i f\ dx^i $$ The left side can only be defined locally, and only with respect to a chosen coordinate system, but because it behaves like a covector under change of coordinates, we could define $df:=\partial_i f\ dx^i$ and still interpret it as a coordinate independent covector.
Problems arise when we try to write a coodrinate independent derivative of tensor fields, though. The coordinate free equivalent of the (nonsense) equation given above would be something like $$ \partial_v T(p):=\lim_{h\to 0}\frac{T(\gamma(h))-T(p)}{h},\ \ \ \ \ \text{$\gamma$ is a path satisfying $\gamma(0)=p$, $\dot{\gamma}(0)=v$} $$ This definition is still nonsense, since it doesn't make sense subtract tensors above different points, and because we want to define things in a coordinate independent way, there's no way to proceed.