(Please note here by just tensors I mean higher order tensors. I know vector is a tensor of rank 1 though).
Consider the following vector:
$\vec v=v^i \vec e_i$
Clearly $v^i$ is not a vector. It is a matrix or array of vector components (more specifically contravariant vector components). It is $\vec v$ that is the vector. $\vec e_i$ is unit vector. So we can see here $\vec v$ & $\vec e_i$ are vectors. Why? because they have both magnitude & direction. You can see the 'vector sign' on v & $e_i$ (unlike $v^i$). $v^i$ is just an array of magnitudes(or scalars).
Lets move on to tensors.
$T= T_{ij}^{k} δ_{k}^{ij}$
$δ_{k}^{ij}$ is the Kronecker delta.
I thought tensors were generalizations of scalar or vector in a complete sense i.e. it applies to both magnitude & direction. When I read this- https://en.wikipedia.org/wiki/Cauchy_stress_tensor , the Cauchy stress tensor (the typical example of a tensor) is an array of only magnitudes; no directions. So doesn't a tensor then have to contain components with direction as well? -I thought. This link too suggests that tensors have components that doesn't have direction- What is the magnitude of a tensor?.
If tensors don't have direction for their components (so basically array of tensor components=tensor unlike for vectors ...& tensor is a generalization of array of vector components; not of vector), wouldn't we need a quantity that would be the generalization of vector per se? (Maybe one can say if vector has a magnitude & a direction, the next immediate higher generalization of vector has magnitude, 1-direction & 2-direction...where 2-direction is the next immediate higher generalization of 1- direction & 1-direction=conventional direction. One may call the new quantity as 2-vector & the conventional vector as 1-vector.And similarly one can do for 3-vector or even n-vector).