Does the definition of the linear span of a subset of a vector space require that the set be countable?
Indeed, $S$ may be an uncountable subset.
However, the notation in any case means that you consider finite sums (not countable sums).
Regardless the cardinality of $S$, the linear span is the set of all elements that you can write as a sum of finitely many terms of the form $\lambda s$ with $\lambda \in \mathbb{R}$ and $s \in S$.
You can use each element of $S$ but you can, for any given sum, only combine it with finitely many other elements of $S$.
For example, if you consider the vector space of infinite real sequences $(x_n)_{n \in \mathbb{N}}$ and for $j \in \mathbb{N}$ you set $e_j$ the sequence that has $j$-th term $1$ and is $0$ otherwise. (So something that resembles the canonical basis in $\mathbb{R}^n$.)
Then the span of $\{e_j \colon j \in \mathbb{N}\}$ is not the full space. Instead the span is the subspaces of sequences that have only a finite number of non-zero terms.
(There is a related notion that does allow countable sums, see Schauder basis but this is something other than what you are considering.)
Let me add that using your notation $v_1$ is not one fixed element of $S$ it is just some kind of dummy-variable. The $v_1$ in one sum, will not be the same as a $v_1$ in another sum (it could be, say, $v_2$ or not appear at all).
The definition says that $S$ is any subset of $V$. The important thing here is to understand that the span of $S$ is the set of all finite linear combinations of elements in $S$.
The condition $k\in\mathbb{N}$ means that the sum (the linear combination) is finite.
The parameter $k$ is the number of vector (in $S$) you take to form the linear combination of elements of S.
To have an element of $L(S)$ you have to:
1) Choose $k\in\mathbb{N}$
2) Choose any $k$ vectors in $S$
3) Form the linear combination of this vectors and this will be your element in $L(S)$. This is, we variate the scalars $\alpha_1,\ldots,\alpha_k$ in $\mathbb{R}$.
To have all elements in $L(S)$ you can vary the parameter $k$ and then you vary the choice of $k$ vectors in $S$ and of course, you vary the scalars $\alpha_1,\ldots,\alpha_k$ in $\mathbb{R}$.
No, restricting $k\in\mathbb{N}$ does not require $S$ to be countable.
Consider the example of $S$ a line through the origin, i.e. the set of $\alpha v$ for some fixed $v$ as $\alpha$ ranges through $\mathbb{R}$. Then $S$ is uncountable, but $L(S)$ is still perfectly well-defined, and indeed $L(S)=S$.