A real vector space with an almost complex structure vs its complexification

Suppose $(V, J)$ is a real vector space of dimension $2n$ equipped with an almost complex structure $J$ such that $J^2 = -Id$. $V$ can be realized as a complex vector space by setting $iv = J(v)$ for any $v \in V$. On the other hand one can consider the complexification of $V$, $V \otimes_{\mathbb R} \mathbb C$ and the complexification can be viewed as a complex vector space naturally. There are two almost complex structures on the complexification, one is the $\mathbb C-$linear extension of $J$ and the other is multiplication by $i$. What is the motivation behind considering the complexification in terms of doing complex geometry? Why can't we just consider the original tangent space?

I understand the complexification can be decomposed into eigenspaces of $J$, but what confuses me a bit is what is the benefit of doing so.

Say $V$ is a real vector space of dimension $2n$, and $J \colon V \to V$ is an operator squaring to $-1$. Then (as you know) the pair $(V, J)$ defines a complex vector space of dimension $n$, with $i \cdot v := Jv$ giving the action of $i$. This means that in any basis, the matrix of $J$ is an $n \times n$ diagonal matrix, with every diagonal entry equal to $i$. Stated in another way, $J$ now satisfies a linear polynomial $J - i = 0$, rather than a quadratic $J^2 + 1 = 0$.

On the other hand, the complexification $V_\mathbb{C} := V \otimes \mathbb{C}$ is also a complex vector space, this time of dimension $2n$, and the action of $i$ is given by $i \cdot (v \otimes z) = v \otimes iz$. The complexification $J_{\mathbb{C}} := J \otimes \operatorname{id}_\mathbb{C}$ is also an operator on this space, sending an element $v \otimes z$ to $Jv \otimes z$. $J_\mathbb{C}$ still satisfies the quadratic $J_\mathbb{C}^2 + 1 = 0$, and hence splits $V_\mathbb{C}$ into two subspaces: the $+i$ eigenspace and the $-i$ eigenspace of $J_\mathbb{C}$. We can see that $$ J_\mathbb{C}(v \otimes 1 - Jv \otimes i) = Jv \otimes 1 + v \otimes i = i(v \otimes 1 - Jv \otimes i),$$ $$ J_\mathbb{C}(v \otimes 1 + Jv \otimes i) = Jv \otimes 1 - v \otimes i = -i(v \otimes 1 + Jv \otimes i),$$ and by further playing around with these eigenspaces you can see that they are each $n$-dimensional. Therefore the matrix of $J_\mathbb{C}$ in an appropriate basis here is diagonal, with $n$ entries equal to $i$, and $n$ entries equal to $-i$.

So in the case of the complex structure $(V, J)$, $J$ simply acts as the $n \times n$ multiplication-by-$i$ scalar operator, while in the complexification $V_\mathbb{C}$, the operator $J_{\mathbb{C}}$ is a particular $2n \times 2n$ diagonalisable operator with eigenvalues $\pm i$ (in particular, it is nothing like a complex structure, but still an operator of a special form).

As for why one wants to consider complex tangent bundles by occasionally forgetting their complex structure and then remembering them again, I would draw an analogy with taking real and complex parts of a complex number (although this analogy is a bit dubious, as a "real structure" on a complex vector space is a different thing...), it allows one to work with all the usual tools one has for real vector spaces or manifolds, then try to use the complex structure to deduce things for the complex manifold.