There are several misconceptions in the OP about both mathematicians' and physicists' use of the word "vector", and even about what scalars and tensors are. To keep this a concise overview I'll be linking to fuller explanations.
Firstly, anything you've heard about magnitude and direction was just an attempt to help schoolchildren avoid certain fallacies without having to explain the entire concept of a vector space to them. The aim is to make sure they understand that, for example, a particle's momentum points a certain way but its amount of energy doesn't.
In general, vectors are not tuples. Admittedly some sets of tuples satisfy the axioms of a vector space if you define arithmetic the usual way, but vectors are so much more general than that case, as examples discussed above show. What is true in general is that, if a vector space $V$ has a basis of the form $\left\{e_i|i\in I \right\}$, then each vector in $V$ is expressible as a linear combination of the $e_i$. Depending on the details, this "linear combination" might be a sum or an integral. Armed with this, the coefficients used can provide a tuple representation of vectors (although in some cases you need infinitely many numbers), but the vector is an independent object. The map is not the territory. In fact, making a terrain look different by creating a new map that's rotated relative to an old one is a special case of what you'll sometimes here called a basis change. Since you're familiar with $\mathbb{R}^n$, I'll give a simple example. The vectors $\left(\begin{array}{c}
1\\
0
\end{array}\right),\,\left(\begin{array}{c}
0\\
1
\end{array}\right)$ comprise a basis of $\mathbb{R}^2$, but I can rotate a 2D map by an angle $\theta$ because $\left(\begin{array}{c}
\cos\theta\\
\sin\theta
\end{array}\right),\,\left(\begin{array}{c}
\sin\theta\\
-\cos\theta
\end{array}\right)$ comprise a basis too.
I should also point out in passing that, while in some contexts the word "basis" simply means a choice of $\left\{e_i|i\in I \right\}$ for which this can be done, the proper definition requires that the linear combination need only use finitely many of the $e_i$. Many vector spaces of interest that do not have finite dimension nonetheless meet some additional technical conditions that allow the less strict meaning of "basis" to be useful. However, the famous statement that two bases of a vector space have the same cardinality refers to the finite-combinations-only definition.
So that's what mathematicians mean by vector spaces. A vector space is always "over" a field of scalars. Just as a vector is defined as an element of a vector space which in turn has a long definition, a scalar is defined as an element of a field which in turn has a long definition.
Or is it? Let's talk about what physicists really mean when they discuss vectors. On the one hand, they know about all the mathematics I mentioned above. On the other hand, they also want to describe nature in terms of quantities that transform in certain convenient ways when we switch coordinate systems, to exemplify "symmetries". This leads them to define "vector" in a stricter way. For example, one thing schoolchildren aren't told is that, although angular momentum has a magnitude and direction, it's not a vector because of the way it transforms under reflections. The distinction in $\mathbb{R}^3$ between vectors and axial vectors takes some explaining. The confusion is understandable. Position and momentum are "in" $\mathbb{R}^3$ and are vectors; angular momentum is "in" $\mathbb{R}^3$ is an axial vector. The reason is simply that none of these things are really "in" a famous set of tuples, because they're not tuples at all; they're quantities that admit a tuple representation. That's one similarity axial vectors have with "true" vectors.
With the development of differential geometry, we realised there is a more elegant way to talk about all this. Instead of distinguishing between true vectors and axial vectors, we can distinguish between contravariant and covariant vectors, provided our "both types count" definition of vector means "rank one tensor". quantity $T^{\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}$ with $p,\,q$ non-negative integers is called a tensor of rank $p+q$ and order (or type) $\left(p,\,q\right)$ iff a coordinate transformation of spacetime from $x^\mu$ to $x^{'\nu}$ obeys $$T^{'\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}=\sum_{\gamma_1\cdots\gamma_p \delta_1\cdots\delta_q}\frac{\partial x^{'\alpha_1}}{\partial x^{\gamma_1}}\cdots\frac{\partial x^{'\alpha_p}}{\partial x^{\gamma_p}}\frac{\partial x^{\delta_1}}{\partial x^{'\beta_1}}\cdots\frac{\partial x^{\delta_q}}{\partial x^{'\beta_q}}T^{\gamma_1\cdots\gamma_p}_{\delta_1\cdots\delta_q}.$$(We never actually write the summation sign; we take for granted that any index that appears twice, once as a subscript and once as a superscript, is summed over all possible values. In relativity, there is one such value for each spacetime dimension.) A tensor of rank $0$ is a scalar, and is unchanged under coordinate transformations. A tensor of positive rank is called covariant if $p=0$, contravariant if $q=0$ and mixed otherwise. Mixed tensors have $p\geq 1$ and $q\geq 1$, so have rank $\geq 2$.
Something that looks like a tensor by virtue of its indices may not transform the right way to actually be a tensor. (Of course, if there are no indices at all something would "look like a scalar", but might not be one.) Here are three important examples.
Best Answer
A single force can be represented by an arrow (a simple assumption)
An arrow is one-dimensional.
As by Google, a
Vector is:
Quantity is:
We know that the length of an arrow represents the power of a force, or in other words, the quantity. Also, we know that the direction of an arrow is it's, well direction, and the magnitude being the quantity. And, also, it is actually determining the position of a point in space relative to another! The comparison is, well an origin on a set of coordinates, longitude and latitude in real life, and there are also many more examples!
From this, we conclude that it is not necessary to classify as something else.
As for your last question, I don't understand what is meant by such quantities, so therefore, I cannot answer that (probably because of my poor comprehension...)