This is a pretty broad question, so I'm not optimistic it'll stay open for long. That said, it's a great observation!
First, I'll answer something you implicitly mentioned before your first question:
What if we don't require scalars to form a field?
If we instead just ask them to form a commutative ring, (i.e. we can do everything you're used to in a field except for division) then the structure we get isn't a vector space any more, but it's a slightly different structure called a module.
There's a lot of interesting things to be said about modules, but the main point is that without division, scaling down isn't generally possible.
This has some surprising differences from a vector space. For instance, in a vector space if you have a spanning set which is not independent, you can throw away useless vectors and make a basis. This isn't always possible in a module, and the notion of 'dimension' is less clearly defined for a module.
For instance, consider the module $\mathbb{Z}$ with the base ring $\mathbb{Z}$. The set $\{2,3\}$ is linearly dependent (in the way that you're used to) and spans $\mathbb{Z}$ since they are coprime, but neither $\{2\}$ nor $\{3\}$ span $\mathbb{Z}$.
In particular, when dealing with vector spaces you'll likely use the fact that if a finite set is linearly dependent, then you can write one of the vectors in terms of the others. How? Well, suppose
$$\sum_{i=1}^n \lambda_i v_i = \lambda_1 v_1 + \dotsb +\lambda_n v_n = 0$$
for $\lambda_i \in \mathbb{F}$ and $v_i \in V$. Since the vectors are linearly dependent, one of the $\lambda_j \neq 0$. Then we can just rearrange to get
$$\lambda_j v_j = -\sum_{i=1\\i\neq j}^n \lambda_i v_i \implies v_j = -\sum_{i=1\\i\neq j}^n \frac{\lambda_i}{\lambda_j} v_i$$
by dividing through by $\lambda_j$. But in a module, we can't divide by scalars. So, such an expression isn't always possible, creating things that can be surprising when you're used to vector spaces.
Is the real number line a vector space?
Yes! In a few ways, actually. First, it's a vector space over itself: take the base field to be the reals, and you get the reals as a one dimensional subspace.
Another way you can make the reals a vector space is to define the base field as the rationals. Then, the real numbers form a vector space over the rationals, and it's not difficult to see it's infinite dimensional. What is difficult however, is to write down a basis, which requires the Axiom of Choice.
Can there be Vector Spaces contained inside the Real Number Line?
Yes! Going in this direction pushes you into Galois theory. For example, taking $\mathbb{Q}$ as the base field again, the vector space with basis $1,\sqrt{2}$ is a ($2$ dimensional) vector space purely contained in the reals (we call it $\mathbb{Q}(\sqrt{2})$. It has elements of the form $a + b \sqrt{2}$, where $a,b$ are rational numbers. Alternatively, you can put $\sqrt[3]{2}$ in your vector space instead, and give the (3 dimensional) vector space with basis $1, \sqrt[3]{2}, \sqrt[3]{2}^2$, also contained in the reals.
You can get even bigger ones by considering the algebraic numbers, which has countably infinite dimension over $\mathbb{Q}$.
You can make the base field larger too, for instance you could use a base field $\mathbb{Q}(\sqrt{2})$, and consider the numbers of the form $a + b \sqrt{3}$, where $a,b \in \mathbb{Q}(\sqrt{2})$, which has dimension 2 over $\mathbb{Q}(\sqrt{2})$ (and dimension 4 over $\mathbb{Q}$).
Finally, as mentioned earlier you can take $\mathbb{R}$ over $\mathbb{Q}$, which has uncountable dimension.
Best Answer
Yes, that's possible. It is not uncommon to consider the complex numbers $\mathbb{C}$ as a two dimensional vector space over the real number $\mathbb{R}$. A basis is given by $1$ and $i$, for example.
All that is needed is that the scalar multiplication is defined in some way. Often this entails that the elements of the vector space are somehow naturally linked to the field over which the space is defined.
However, I can say $\{q,u,i,d\}$ is a vector space over the field $\mathbb{Z}/2 \mathbb{Z}$, with $q+ x = x $ for each $x \in \{q,u,i,d\}$, further $u+u = i+i = d + d = q$, and $u+i = d$, $u+d = i$, $i+d = u$. Further $\overline{0} x = q$ and $\overline{1} x = x$ for each $x$ in $\{q,u,i,d\}$.
A more natural example is proposed by Rob Arthan: For any set whatsoever $X$ the powerset $\Bbb{P}(X)$ becomes a vector space over $\Bbb{Z}/2\Bbb{Z}$ if one defines the sum of two sets to be their symmetric difference so that $0 = \emptyset$ and the action of $\Bbb{Z}/2\Bbb{Z}$ in the only possible one: $0A = \emptyset$ and $1A = A$.