Given a field $K$ and $V$ (is a vectorial K-space). Additive group of field $K$ acts on additive abelian group $(V,+)$ with a scalar multiplication $K\times V \rightarrow V$ such that $(\lambda,v)\rightarrow \lambda v$.
For each $v \in V$ the orbit is $orb_v(k)=\{a\in K:k\sim a\}$ then $\{a \in K:a=kv$ for some $k \in K \}$ then $orb_v(k)=$ right coset of $g$
Am I ok?
What you've written doesn't make much sense to me. First of all, the map:
$(k,v) \mapsto k \cdot v = kv$ is not an action of $(K,+)$ on $(V,+)$, for if it were, we would need to have:
$0\cdot v = v$, for all $v \in V$, but $0\cdot v = 0_V$ for all $v \in V$ (including non-zero vectors $v$, if there be any).
It is however, an action of $(K \setminus \{0\},\ast)$ on $(V,+$), since:
$k_1\cdot(k_2\cdot v) = k_1\cdot (k_2v) = k_1(k_2v) = (k_1k_2)v = (k_1k_2)\cdot v$, by the vector space axioms, and:
$1_K \cdot v = 1_Kv = v$, again, by the vector space axioms.
The orbit of any $v \in V$ in this action is any (and all) scalar multiples of $v$, that is to say, the line through $0_V$ and $v$.
For $K = \Bbb R$, and $V = \Bbb R^2$, the orbit space may be identified with $\Bbb {RP}^1$, the real projective line, the "points" on the line are the orbits.