It's helpful to think like a computer scientist here...
a tensor is a type of multidimensional array with certain transformation properties
So if we wrote the spec for array we just have to specify the collection of numbers.
array(N1, N2, N3)
If we specify tensor, we need to specify how this array "transforms" under certain matrix operations.
A spinor is going to be a tensor with a group action in addition to the other transformation rules. In particular an $SU(2)$ group action.
This already happens in linear algebra. Take the function $T: (1,0) \mapsto (1,1), (0,1) \mapsto (1,-1)$. We can write this as a $2 \times 2$ matrix:
$$\left[ \begin{array}{cr} 1 & 1 \\ 1 & -1 \end{array}\right]$$
These matrices have eigenvalues and eigenvectors. We have to solve the equation:
$$ \lambda^2 + 2 = 0$$
We get eigenvectors of $\lambda = \pm \sqrt{2}$. The eigenvectors are $(1 \pm \sqrt{2}, 1) $ and the diagonal form is:
$$\left[ \begin{array}{cr} \sqrt{2} & 0 \\ 0 & -\sqrt{2} \end{array}\right]$$
The same geometric object a "linear transformation" can be one of two different matrices depending on the basis we choose!
In differential geometry we can consider the metric tensor:
$$ dz^2 = dx^2 + dy^2 $$
If we switch to polar coordinates: $x = r \cos \theta, y = r \sin \theta$. Then $dx = -r \sin \theta \, d\theta + \cos \theta \, dr$ and $dy = r \cos \theta \, d\theta + \sin \theta \, dr$. Then
\begin{eqnarray*} dz^2 &=& (-r \sin \theta \, d\theta + \cos \theta \, dr)^2
+ ( r \cos \theta \, d\theta + \sin \theta \, dr)^2 \\
&=& (r d\theta)^2 + dr^2
\end{eqnarray*}
It should not matter if we compute the arc length in polar coordinates or cartesian, it should give the same answer.
