A function $f : Z \rightarrow W$ over a field $\mathbf F$, where $Z,W$ are vector spaces in $\mathbf K$, is typically called linear if for all $a, b \in F$ and all scalars $c \in \mathbf F$, we have
$f(\mathbf a + \mathbf b) = f(\mathbf a) + f(\mathbf b),$
$f(c \mathbf a) = c f(\mathbf a).$
One thing I noticed is that the second condition is not necessary if we restrict $c$ to be rational. See the lemma below.
Lemma. If $c$ is rational, then $f(\mathbf a + \mathbf b) = f(\mathbf a) + f(\mathbf b)$ implies $f(c \mathbf a) = c f(\mathbf a).$
Proof. Let $c = m/n$ where $m,n \in \mathbb Z$ and define $\mathbf z = \mathbf a/n$. We thus have
$$f(c\mathbf a) = f(m\mathbf a / n ) = f(m\cdot \mathbf z) = f(\underbrace{\mathbf z + \mathbf z + \cdots + \mathbf z}_\text{$m$ times}) = \underbrace{f(\mathbf z) + f(\mathbf z) + \cdots + f(\mathbf z)}_\text{$m$ times} = m f(\mathbf z).$$
On the other hand, $$f(\mathbf a)= f(n \mathbf z) = f(\underbrace{\mathbf z + \mathbf z + \cdots + \mathbf z}_\text{$n$ times}) = \underbrace{f(\mathbf z) + f(\mathbf z) + \cdots + f(\mathbf z)}_\text{$n$ times} = n f(\mathbf z),$$ $$ f(\mathbf z) = \frac{1}{n} f(\mathbf a).$$
Therefore, $$ f(c\mathbf a) = m f(\mathbf z) = \frac{m}{n} f(\mathbf z) \Rightarrow \boxed{ f(c\mathbf a) = cf(\mathbf z) }$$
Can this idea be extended in general for all scalars $c$? can we remove the condition $f(c \mathbf a) = c f(\mathbf a)$ and still have the same definition of a linear function? If not, what would be a counterexample of a function $f$ that satisfies $f(\mathbf a + \mathbf b) = f(\mathbf a) + f(\mathbf b),$ but not $f(c \mathbf a) = c f(\mathbf a)$?
No, it can't be extended beyond $\mathbb{Q}$, or another prime field $\mathbb{Z}/p$. (if our field has finite characteristic)
Consider complex conjugation as a map from the $\mathbb{C}$-vector space $\mathbb{C}$ to itself. Is this linear? The first condition holds; $\overline{a+b}=\overline{a}+\overline{b}$. The second doesn't; $\overline{ia}=-i\overline{a}\neq i\overline{a}$.
Every field $K$ that isn't $\mathbb{Q}$ or some $\mathbb{Z}/p$ can be made into a vector space over one of those fields, in a natural way - and the copy of that field contains all of the multiples of $1$. Consider a map from $K$ to itself, linear as a vector space map over the prime field. Unless this map is a multiple of the identity, it won't be linear as a map of the $K$-vector space $K$ to itself - and there are too many linear maps to all be multiples of the identity. Such a map will distribute over addition, but not commute with scalar multiplication.