Let $ V $ an $ \mathbb{R} $-vector space, we can prove that $ V $ is a manifold. Using the topology compatible with $ \mathbb {R} ^ n $ and defining the chart $ (V, \phi_B) $ where $ \phi_B: V \to \mathbb {R} ^ n $, $\phi_B(v)=(\lambda_1,...,\lambda_n)$, $B=\{v_1,...,v_n\}$ is a base of $V$ and $v=\lambda_1 v_1+...+ \lambda_nv_n$. I want to show that this structure does not depend on the base, that is, for another base $B'=\{v_1', ..., v_n'\}$ when we define $\phi_{B'}$ we have the compatibility of the charts $\phi$ and $\phi'$, i.e, $\phi \circ \phi'^{-1}$ and $\phi' \circ \phi^{-1}$ are smooth.
To find the composition explicitly I thought of introducing the base change, but I can't conclude that the compositions are smooth.
Note that the charts you define are all globally defined, that is, they have domain $V$. Hence to prove that change of coordinates are smooth is suffices to compute
$$ \varphi_B \circ \varphi_{B'}^{-1} : \mathbb{R}^n \to \mathbb{R}^n $$
and prove it is smooth (the other composition is automatic by interchanging $B$ with $B'$).
As you say, this is a base change. But a simpler observation suffices: this map is a composition of linear ones, hence linear itself, and so it is smooth.