A vector field can contain sinks. But a tensor field does not. Why is it so?
(I am referring to the statement below from this paper)
"A degenerate point of a tensorfield is equivalent to a singularity in a vector field. However,a vector field usually contains sinks which can trap the robot and more over, can cause oscillating movement around them due to numerical stability issue.In contrast,a tensor field contains only two types of degenerate points (Figure5), namely wedges and trisectors,but not sinks,thus avoiding the local trapping issue."
Tensors are often used to describe ordering when directionality can fail you. I will use liquid crystals as an example because the theory of Q-tensors used there a lot.
Imagine the gas (liquid) of arrows. And you want to describe its ordering. What you can do is you can assign a normalized vector to each arrow $|\vec n|=1$ and take the average $\vec p=\left< \vec n\right>$. So now you are able to distinguish chaotic arrangment of arrows vs strictly aligned one:
The left part of image will give you $\vec p=0$ and the right one will give you $\vec p=(1,1)/\sqrt2$.
But if you apply the same method to two pars of this image, both parts will give you zero.
However, it's clear that right one has some sort of alignment, whereas the left one is still chaotic. The situation can get worse, if there were no arrows in the first place, but some symmetric sticks (like toothpicks with two sharp ends).
So instead of just direction vector $\vec n$, you use orientation tensor $q=\vec n\otimes\vec n$. In 2d it can be written as: $$ q = \begin{pmatrix} n_x^2 & n_xn_y \\ n_xn_y& n_y^2\end{pmatrix}. $$
(Usually, one also would like to make it traceless $q = \vec n\otimes\vec n - \frac 1NI$, but it doesn't change much in our case)
So now we can average q-tensors: $Q=\left<q\right>$ and distinguish between those 2 cases. See, how change of sign $\vec n\to-\vec n$ keeps $q$ the same. So q-tensors are natural choice when you want to talk about orientation but not direction.