If $v$ is a vector field in $\mathbb{R}^{2n}$, $J$ is an almost complex structure and $f\colon \mathbb{R}^{2n}\longrightarrow \mathbb{C}$ such that $f=f_{1}+if_{2}$, we can define $v\cdot f= (v\cdot f_{1})+i(v\cdot f_{2})$. Then, $f\mapsto v\cdot f$ is a complex linear map.
Being complex linear means that if $\varphi$ is the map that sends $f$ to $v\cdot f$, then $\varphi \circ J=J \circ \varphi$.
What I do not understand is how is $J$ defined on maps from $\mathbb{R}^{2n}$ to $\mathbb{C}$. I mean, $J_{p}\colon T_{p}(\mathbb{R}^{2n})\longrightarrow T_{p}(\mathbb{R}^{2n})$, so $J_{p}$ has to be evaluated in elements of the tangent space.
What am I understanding wrong?
Let's be more explicit. Your definitions in fact don't require any complex structure. Let $U$ be a real vector space (considered also as a smooth real manifold) and denote by $C^{\infty}(U,\mathbb{C})$ the space of smooth complex-valued functions $f \colon U \rightarrow \mathbb{C}$ on $U$. Any such function can be written uniquely as $f = f_1 + i f_2$ where $f_1,f_2 \colon U \rightarrow \mathbb{R}$ are smooth real-valued functions. The space $C^{\infty}(U,\mathbb{C})$ has a natural structure of a complex vector space. Explicitly, if $z = a + ib \in \mathbb{C}$, we define
$$ (f + g)(x) := f(x) + g(x) = (f_1(x) + i f_2(x)) + (g_1(x) + i g_2(x)) = (f_1(x) + g_1(x)) + i (f_2(x) + g_2(x)), \\ (zf)(x) := (a + ib)(f_1(x) + i f_2(x)) = (af_1(x) - bf_2(x)) + i (af_2(x) + bf_1(x)). $$
Now, given any vector field $v$ on $U$, it acts on smooth real-valued functions in the usual way. You purpose to extend this definition so that
$$ (vf)(x) := ((vf_1)(x) + i(v f_2)(x)). $$
Thus, any vector field $v$ on $U$ gives you a map $v \colon C^{\infty}(U,\mathbb{C}) \rightarrow C^{\infty}(U,\mathbb{C})$ and this map is $\mathbb{C}$-linear in the sense that $v(zf) = zv(f)$ for all $z \in \mathbb{C}$ and $f \in C^{\infty}(U,\mathbb{C})$ (and of course $v(f+g) = v(f) + v(g)$).