Vector Components Under Coordinate Transformations-[conceptual understanding] I have picked up Tensor Calculus very recently, having covered 3-part sequence of undergraduate calculus in school, Linear Algebra and a Math Methods course for Physicists I felt I had the necessary background to start understanding Tensors; but there is this one thing I still cannot understand properly.
In Dwight E. Neuenschwander's Tensor Calc. for Physics equations (1.69)&(1.70), he writes that in transforming from a coordinate system $x^j$ to ${x^i}'$, the vectors are objects whose components transform according to: $$ {A'}^i=\sum_j \frac{\partial{x'}^i}{\partial{x}^j}A^j \qquad\qquad(1)$$  because displacements transform under the transformation as : $$ {dx'}^i=\sum_j \frac{\partial{x'}^i}{\partial{x}^j}dx^j\qquad\qquad(2)$$
I have a hard time comprehending this definition of vectors, before this we have had exposure to component transformation under a rotation of axes in Analytical Geometry course but how are the two related? I understand we have vectors thought of as objects with components transforming in specific ways when we change coordinates, but I am not sure how we are relating vector components to dx here.
Further questions i have are as follows:
[1] in equations (1)&(2) what are the indices i and j? why are we only summing over j here?
[2]Dwight only writes the transformation law, but how would i actually apply this?, assuming i have a vector A in cartesian coordinates and I switch to polar coordinates, how would i really use this law?-an example would help a lot.
[3] IN some questions I have seen that in testing for vector character of certain quantitites we only test under a rotation of axes, why couldnt we have opted for this definition then?.
 A: When To Sum Indices
In tensor calculus, changes in vector components due to coordinate transformations are always computed as a linear combination of the derivatives of the new coordinate component $x'^i$ with respect to each coordinate component from its preceding old system $x^j$. This stems from the fact that coordinate conversion equations are usually functions of multiple components from an old coordinate system. According to the Einstein summation convention, indices in differential geometry/linear algebra equations must be summed over each dimensional coordinate if and only if they are repeated within a term (usually in the form of one upper/lower index). Such indices are referred to as dummy indices. This summation convention exists for the sake of compactness since, unsurprisingly, nearly all tensor formulae are expressed as combinations of coordinate components.
Example: line element expansion using the metric tensor $g_{\mu\nu}$: $$ds^2=g_{\mu\nu}dx^\mu dx^\nu=-g_{tt}c^2dt^2+g_{rr}dr^2+g_{\theta\theta}d\theta^2+g_{\phi\phi}d\phi^2$$ The Einstein summation convention is implied for $\mu$ and $\nu$, rendering $\sum$ redundant.
In the case of the following transformation equation:
$$dx'^i=\sum_{j=1}^{n}\frac{\partial x'^i}{\partial x^j}dx^j$$
whose $\sum_j$ can be dropped...
$$dx'^i=\frac{\partial x'^i}{\partial x^j}dx^j\qquad(2)$$
it is implied that $dx^j=\{dx^0, dx^1, dx^2, ..., dx^n\}$ be summed over due to $j$ being a dummy index. On the other hand, the free index $i$ is not summed over since it does not repeat itself. Therefore, $i$ can represent any coordinate component we choose. Our result is the following expansion:
$$dx'^i=\frac{\partial x'^i}{\partial x^1}dx^1+\frac{\partial x'^i}{\partial x^2}dx^2+...+\frac{\partial x'^i}{\partial x^n}dx^n$$
Note: greek indices such as $\mu$ or $\nu$ imply the inclusion of a time component in a summation whereas Latin indices such as $i$ or $j$ merely imply a summation of spatial components.
Example in Practice
Consider the transformation of a vector $\overrightarrow{A}$ from cartesian to polar coordinates whose components (as a linear combination) are:
$$\overrightarrow{A}=3\overrightarrow{e_x}+5\overrightarrow{e_y}+4\overrightarrow{e_z}$$
The conversion relations from $(x,y,z)\rightarrow(r,\theta,\phi)$ read:
$$r=\sqrt{x^2+y^2+z^2},\qquad\theta=\arctan\bigg(\frac{y}{x}\bigg),\qquad\phi=\arctan\bigg(\frac{\sqrt{x^2+y^2}}{z}\bigg)$$
Using eq. (2), we can plug each conversion relation into each partial derivative term.
$$dr=\frac{\partial r}{\partial x}dx+\frac{\partial r}{\partial y}dy+\frac{\partial r}{\partial z}dz$$
$$=\frac{\partial}{\partial x}\bigg(\sqrt{x^2+y^2+z^2}\bigg)dx+\frac{\partial}{\partial y}\bigg(\sqrt{x^2+y^2+z^2}\bigg)dy+\frac{\partial}{\partial z}\bigg(\sqrt{x^2+y^2+z^2}\bigg)dz$$
$$=\frac{x}{\sqrt{x^2+y^2+z^2}}dx+\frac{y}{\sqrt{x^2+y^2+z^2}}dy+\frac{z}{\sqrt{x^2+y^2+z^2}}dz$$
Doing the same with $d\theta$ and $d\phi$, we end up with:
$$d\theta=-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$
$$d\phi=\frac{xz}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}dx+\frac{yz}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}dy-\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2}dz$$
For the sake of accessibility, lets arrange each $\frac{\partial x'^i}{\partial x^j}$ term in a Jacobian matrix:
$$\pmb{J}=\begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} & \frac{\partial r}{\partial z} \\ \frac{\partial \theta}{\partial x}&\frac{\partial \theta}{\partial y}&\frac{\partial \theta}{\partial z} \\ \frac{\partial \phi}{\partial x}&\frac{\partial \phi}{\partial y}&\frac{\partial \phi}{\partial z} \end{bmatrix}=\begin{bmatrix} \frac{x}{\sqrt{x^2+y^2+z^2}} & \frac{y}{\sqrt{x^2+y^2+z^2}} & \frac{z}{\sqrt{x^2+y^2+z^2}} \\ -\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}& 0 \\ \frac{xz}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}&\frac{yz}{\sqrt{x^2+y^2}(x^2+y^2+z^2)}&-\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2} \end{bmatrix}$$
The last step from here is simply to substitute each entry in the Jacobian matrix into the vector transformation eq. (1), with $(x,y,z)\rightarrow(3,5,4)$ as depicted in the aforementioned vector above.
$$A'^i=\frac{\partial x'^i}{\partial x^j}A^j\qquad(1)$$
$$A^r=\frac{\partial r}{\partial x}A^x+\frac{\partial r}{\partial y}A^y+\frac{\partial r}{\partial z}A^z$$
$$=\frac{x}{\sqrt{x^2+y^2+z^2}}A^x+\frac{y}{\sqrt{x^2+y^2+z^2}}A^y+\frac{z}{\sqrt{x^2+y^2+z^2}}A^z$$
$$=\frac{3}{\sqrt{3^2+5^2+4^2}}(3)+\frac{5}{\sqrt{3^2+5^2+4^2}}(5)+\frac{4}{\sqrt{3^2+5^2+4^2}}(4)=5\sqrt{2}$$
I'll leave $A^\theta$ and $A^\phi$ as exercises but they both follow the same procedure as $A^r$.
Why Use This Method Instead of Matrix Rotations?
Methods like eq. (2) generalize coordinate transformations as much as possible for any circumstance, extending beyond merely rotating vectors or coordinates. Coordinate transformations are used all the time in astrophysics and general relativity where physicists create arbitrary new coordinate systems to write tensors (usually spacetime metrics) in a more compact and physically intuitive manner.
Additionally, keep in mind that the variables used in tensor components are not always restricted to dimensional coordinate components. For example, the Kerr-Newman metric tensor (general relativity) describes the curvature of spacetime around a spherically symmetric, charged, rotating mass. In such circumstances, dimensional coordinates alone are simply not a sufficient description since the tensor must also introduce angular momentum and charge as influential geometric factors.
