Why can a vector from an infinite-dimensional vector space be written as finite linear combination?

For a general vector space $V$ it doesn't make sense to talk about infinite sums. I suppose you could define a norm on $V$, as it's a vector space over $\mathbb R$ or $\mathbb C$, but this doesn't generalize to other fields, such as $\mathbb Z / 2$. When it comes to general vector spaces, you can really only talk about finite sums. You can talk about infinite sums in, say, a Hilbert space, but that's a lot more structure.

The very definition of a basis $B \subseteq V$ is that every element in $V$ is a unique finite linear combination of elements in $V$, even if $B$ is infinite. For example, take $V = \{a \in \mathbb R^\mathbb N : a_n = 0 \text{ for all but finitely many } n\}$. Then letting $e_i(j)$ be 1 for $i = j$ and 0 otherwise, we have that $\{e_0, e_1, e_2, e_3, \dots\}$ is a basis for $V$. Although this is infinite, every element of $V$ is a finite linear combination of these basis elements. However, this set is not a basis for $\mathbb R^\mathbb N$. Indeed, the sequence $(1, 1, 1, 1, \dots)$ is not in the span. However, it is a theorem that all vector spaces have a basis, so there is a way to represent all of these sequences as a unique finite linear combination of other sequences. I can't write this basis down for you, as this theorem uses the axiom of choice (and is, in fact, equivalent to it). So if you accept the axiom of choice, your problem can be remedied by just saying "take some basis" without worrying about what it is. If you don't, then there will be some infinite dimensional spaces which don't admit a basis, so you can't always represent vectors in this way.

If all we know is that $V$ is a vector space, then "an infinite sum of vectors" is not necessarily defined.

There are some contexts where infinite sums are defined. For instance, if $V$ has a norm, then we have a notion of distances between vectors, and we can say that $\sum_{n=1}^\infty v_n = v$ if $$ \lim_{N \to \infty} \left\| v - \sum_{n=1}^N v_n\right\| = 0. $$ If $\{v_n\}$ is a "basis" in the sense that every $v$ can be written in the form $v = \sum_{n=1}^\infty a_nv_n$ (for some coefficients $a_n$), then we say that $\{v_n\}$ is a Schauder basis. By contrast, the only kind of basis that makes sense without something like our additional norm structure (i.e. for an arbitrary vector space) is a Hamel basis.