Take, for example $V = \mathbb R ^2$, the $x$-$y$ plane. Write the vectors as coordinates, like $(3,4)$.
Such a coordinate could be written as a sum of its $x$ component and $y$ component: $$(3,4) = (3,0) + (0,4)$$
and it could be decomposed even further and written in terms of a "unit" x vector and a "unit" y vector:
$$(3,4) = 3\cdot(1,0) + 4\cdot(0,1).$$
The pair $\{(1,0),(0,1)\}$ of vectors span $\mathbb R^2$ because ANY vector can be decomposed this way:
$$(a,b) = a(1,0) + b(0,1)$$
or equivalently, the expressions of the form $a(1,0) + b(0,1)$ fill the space $\mathbb R^2$.
It turns out the $(1,0)$ and $(0,1)$ are not the only vectors for which this is true. For example if we take $(1,1)$ and $(0,1)$ we can still write any vector:
$$(3,4) = 3\cdot(1,1) + 1\cdot(0,1)$$
and more generally
$$(a,b) = a \cdot (1,1) + (b-a)\cdot(0,1).$$
This fact is intimately linked to the matrix $$\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$$ whose row space and column space are both two dimensional.
I would keep going, but your question is general enough that I could write down an entire linear algebra course. Hopefully this gets you started.
You need not choose vectors from the canonical basis for $\mathbb{R}^4$; any basis will do. There may be more convenient bases from which you can add vectors, if it's easier to prove that those vectors are linearly independent with the vectors you already have.
Using the canonical basis is certainly not wrong (and it is an intuitive choice), but you should know that there is no property unique to this basis that makes this process work.
Best Answer
Indeed they span different subspaces. Why would you think that they span the same subspace?
Consider a simpler example that can be visualised easily. Consider $R^3$ and clearly $(1,0,0)$ and $(0,1,0)$ span the entire $xy-$plane however, $(0,1,0)$ and $(0,0,1)$ span the entire $yz-$ plane.
Why so? It is because taking only a few basis doesn't exhaust all the cases. In this case, taking only two such as $(1,0,0)$ and $(0,1,0)$ doesn't consider the case $(0,1,0)$ and $(0,0,1)$.
To generate a subspace, no. of linearly independent vectors should be equal to no. of basis of that subspace which is also called dimension of the subspace.