A vector space is defined as a quadruple $(\mathbf{V},\mathbb{K},\oplus,\odot)$ where $\mathbf{V}$ is a set of elements called vectors, $\mathbb{K}$ is a field $(\mathbb{K},+,\cdot)$ (and we say that the vector space is a space over $\mathbb{K}$) ,
$\oplus$ is a binary operation (called sum) on $\mathbf{V}$ such that $(\mathbf{V},\oplus)$ is a commutative group and
$\odot:\mathbb{K}\times\mathbf{V} \rightarrow \mathbf{V}$ is a scalar multiplication such that, $\forall a,b \in \mathbb{K}$ and $\forall \mathbf{u,v} \in \mathbf{V}$ we have:
$$
a\odot(b\cdot\mathbf{v})=(a\cdot b)\odot\mathbf{v}
$$
$$
1\odot\mathbf{v}=\mathbf{v}
$$
$$
a \odot (\mathbf{u}\oplus\mathbf{v})=a \odot\mathbf{u}\oplus a\odot \mathbf{v}
$$
$$
(a+b)\odot \mathbf{v}=a\odot \mathbf{v}\oplus b\odot \mathbf{v}
$$
Note that $(\oplus, \odot)$ are, in genral, different from the operations $(+,\cdot)$ in $\mathbb{K}$ .
Now look at your definition:
If $S$ is any non-empty set, then
$V ={f : S\to F}$ ($F$ denotes field.)
is a vector space over the field F with the usual operations of “addition” and “multiplication by a
scalar” of functions.
This define:
The set $\mathbf{V}$ is the set of functions from $S$ to a field $F$.
The field $\mathbb{K}$ is the same field $F$ and the two operations are:
$$
\oplus : \mathbf{V}\times \mathbf{V} \to \mathbf{V} \quad (f+g)(x)=f(x)+g(x) \quad \forall x \in S.
$$
that is: the sum of the two functions is the function that has as value the sum of the values of the two functions.
$$
\odot:\mathbb{K}\times\mathbf{V} \rightarrow \mathbf{V} \qquad (cf)(x)=c(f(x))
$$ that is: the product of a function $f$ with a scalar is the function that has as values the products of the values of the function for the same scalar.
It is not difficult to prove that with these definition all the axioms are satisfied and we have a vector space.
The other examples in your question are similar. E.g: , it is simple to prove that any field $\mathbb{F}$ is a vector space over itself simply defining $\mathbf{V}=F$, $\mathbb{K}=\mathbb{F}$ and $\oplus=+$ and $\odot=\cdot$.
Finally, note that $\mathbb{C}$ is a vector space ( of dimension 2) over $\mathbb{R}$ because a complex number $ x+iy$ can be identified with the couple of real numbers $(x,y) \in \mathbb{R}^2$ and $\mathbb{R}^2$ is a vector space over $\mathbb{R}$ with the usual operations.
