For example, let $V$ be a complex vector space (vector space over C) of dimension 2 and $W=V\oplus V.$
Question:
1) Let $V^R$ denotes the decomplexification of V. Is W the complexification of $V^R$? I don't think so, because the complexification of $V^R$ should be given by the direct sum of real vector spaces. So, in this case, the complexification of $V^R$ should be $(V^R)^C = V^R \oplus V^R $.
Edit: I forgot to include that I am supposing $(V^R)^C$ has a linear complex structure that allows complex scalar multiplication. The complex structure is defined by the operator $J$ that is defined by $J(v,w)=(-w,v)$ for any $(v,w)$ element of $(V^R)^C$. So, for $(a+ib)$ in C, $(a+ib)(v,w)=(a+bJ)(v,w) = (av-bw, aw+bv)$.
2) If $W$ is not the complexification of $V^R$...
$W$ is clearly a complex vector space of dimension $4$. $(V^R)^C$ is a complex vector space of dimension 4 as well (since it has the same dimension as $V^R$). Are $W$ and $(V^R)^C$ isomorphic? if not, what is the relation between them?
I would appreciate any guidance. Thank you.
I think there is a misconception going on here. Taking the direct sum of a real vector space with itself $V \oplus V$ is still a vector space over $\mathbb R$. It is true that it has twice as many basis elements and can be thought of as $\mathbb R^{2n}$, but there is still no way to multiply complex numbers here, since scalar multiplication is not by complex numbers.
What you are looking for is $V \otimes \mathbb C$, which is a tensor product, and as a real vector space, it certainly has dimension $2n$ with basis "elements" $\{e_1, \dots e_n,ie_1, \dots ie_n\}$, but this is a misleading way of writing $e_i \otimes 1$ and $e_i \otimes i$, and the previous terminology is an abuse of notation.
From this point of view, we obtain a complex vector space by scalar multiplication defined by $c \times (v \otimes (a+bi)=v \otimes c(a +bi)$ with $c \in \mathbb C$.
In general, there is an identification $V \oplus i V \cong V \otimes \mathbb C=V^{\mathbb C}$,
but to answer your question: no, just taking a direct sum is not by itself a complexification.
Your last paragraph is also incorrect: $\mathrm \dim_{ \mathbb C}V^{\mathbb C}=\mathrm{dim}_{\mathbb R} V$, which is because every element can be written as a linear combination of complex numbers multiplied by vectors in the usual basis.