Constructivist defininition of linear subspaces of $\mathbb{Q}^n$?
Your question almost answers itself, but this may not be so obvious. So I will try to give a short and clear answer.
Regarding 1) of your question:
As a mathematician well-schooled in intuitionistic and constructive mathematics, I would always consider more than one constructive notion of 'linear subspace'.
Your first definition is definitely my favourite for the purest definition. So I would say: $(A,+,\cdot, k)$ is a linear subspace of $(V, +,\cdot, k)$ iff $A\subseteq V$ is closed under addition and scalar multiplication.
Important extra features would be to say that $(A,+,\cdot, k)$ is a finitely generated linear subspace iff there are $a_1,\ldots,a_n$ in $A$ such that $A=\mathrm{span}\{a_1,\ldots,a_m\}$, and a finite-dimensional linear subspace iff there is a basis $(a_1,\ldots,a_m)$ of $A$.
For $\mathbb{Q}^n$, any finitely generated linear subspace is a finite-dimensional linear subspace, since the equality of elements is decidable on $\mathbb{Q}^n$. But this does not hold for $\mathbb{R}^n$, and so for $\mathbb{R}^n$ we can have a finitely generated subspace which is not provably finite-dimensional. Brouwer gave many examples of such situations.
Regarding 2) of your question:
In intuitionistic mathematics one can easily prove that not all linear subspaces of $\mathbb{Q}^n$ are finitely generated.
Regarding 3) of your question:
Yes, there is a reason to favour your first definition over the second, because in many constructive situations we do not explicitly have a finite set of generators, and still we can do a lot of worthwhile stuff with an $A$ which is closed under addition and scalar multiplication.
[update to reflect the comments below:]
As an indication of constructive situations where linear subspaces of $\mathbb{Q}^2$ which are 'not' finitely generated occur, let me give the following example (a bit contrived to keep it simple, but in higher math these situations are commonplace).
For $\alpha$ in $\{0,1\}^{\mathbb{N}}$ define $V_{\alpha}=\{(0,0)\}\cup\{r\cdot (0,1)\mid r \in\mathbb{Q}\mid \exists n\in\mathbb{N}[\alpha(n)=1]\}$.
Now for an arbitrary $\alpha$ the assertion that $V_{\alpha}$ is finitely generated implies that $\alpha=\underline{0}$ or $\alpha\neq\underline{0}$. We cannot constructively assert the latter for all $\alpha$ in $\{0,1\}^{\mathbb{N}}$, so if we want to prove something for all $V_{\alpha}$ we are stuck with 'non-finitely generated' linear subspaces of $\mathbb{Q}^2$. And in many situations like this often only the closedness of $V_{\alpha}$ under addition and scalar multiplication is necessary to produce a nice general theorem.
To answer question 2 (although perhaps not under a “system of axioms”), here is a simpleminded example showing that if the definitions are equivalent (even just for $n=1$), then $p\lor\neg p$ is $\top$ for any truth value $p$, i.e. the Law of Excluded Middle holds. Given a truth value $p$ consider the subspace $\mathscr{V} = \{r\in\mathbb{Q} : (r=0) \lor p \} = \{0\} \cup \{r\in\mathbb{Q} : p\}$. Clearly this contains zero and is closed under addition and multiplication by $\mathbb{Q}$. But if $\mathscr{V}$ is the span of of a finite family $r_1,\ldots,r_m$ of elements, using the fact that each element of $\mathbb{Q}$ satisfies $(r=0) \lor\neg (r=0)$ (and proceeding by induction on $m$), either every element of the family is zero (in which case $\neg p$) or there is one which is nonzero (in which case it is invertible so $\mathscr{V}$ contains $1$, so $p$ is true). So we have $p\lor\neg p$.
To make this perhaps more transparent, here is a sheaf model of this situation: let $X$ be a topological space, $U \subseteq X$ be an open set such that the union of $U$ and of the largest open set $\neg U$ disjoint with $U$ is not $X$. And consider the subsheaf $\mathscr{V}$ of the constant sheaf $\mathbb{Q}_X$ (sheaf of locally constant $\mathbb{Q}$-valued functions) consisting of those locally constant $r\colon W\to\mathbb{Q}$ (for varying open $W\subseteq X$) which are identically zero outside of $U$. This is a $\mathbb{Q}_X$-module subsheaf of $\mathbb{Q}_X$ which cannot be spanned by global sections.
Note that while my example is not spanned by a finite set $r_1,\ldots,r_m$, there is a “subfinite”(?) set $\{1 : p\}$ (meaning $\{r\in\mathbb{Q} : (r=1) \land p\}$) which spans $\mathscr{V}$ (because each element of $\mathscr{V}$, either is zero in which case it is the linear combination of the empty set, or is not zero, in which case $p$, in which case the element is the linear combination of $1$ with itself as coefficient). I guess this set is even a basis of $\mathscr{V}$, though I'm not sure there aren't many subtly inequivalent definitions.