We define $\mathbb{C}$ to be the triple $(\mathbb{R}^2,+,\cdot)$ where $+$ is the standard componentwise addition and $\cdot$ is defined by $$(x_1,y_1) \cdot (x_2,y_2) := (x_1x_2 - y_1y_2, x_1y_2 + y_1x_2)$$
We have a field embedding $\mathbb{R} \hookrightarrow \mathbb{C}$ by $x \mapsto (x,0)$. Therefore we can "identify" the real number $x$ with $(x,0)$.
$\mathbb{C}$ is a $\mathbb{R}$ vector space.
I mean the vector addition is simply given by addition in $\mathbb{C}$. However, we have to define scalar multiplication by elements of $\mathbb{R}$. I mean it is obvious that we define $\cdot : \mathbb{R} \times \mathbb{C} \to \mathbb{C}$ by $\cdot(\lambda,z) = \lambda z$ where the righthandside means multiplication in $\mathbb{C}$. Now my question is, to be completely rigorous, we should define the scalar multiplication by $$\cdot(\lambda,z) = \iota(\lambda)z$$ where $\iota$ is the embedding of $\mathbb{R}$ into $\mathbb{C}$. But this is somehow never done, instead we simply consider the scalar multiplication $\cdot: \iota(\mathbb{R}) \times \mathbb{C} \to \mathbb{C}$, I think.
There is no problem here. For any $\lambda\in\Bbb R$ and $(x,y)\in\Bbb C$, we have
$$\iota(\lambda)\cdot(x,y)=(\lambda,0)\cdot(x,y)=(\lambda x,\lambda y)=\lambda(x,y),$$
so the "rigorous" way of defining the scalar multiplication agrees with the "naive" way.
Also, I may be misunderstanding your question so let me know if that's the case.