Let's consider the first example. The vector space $W$ can be described as the solutions of this system of linear equations:
$$\underbrace{\begin{bmatrix} 1 & -2 & 1 \\ 2 & -3 & 1 \end{bmatrix}}_{=:A} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
By elementary row operations we get $A$ into the form
$$ B = \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \end{bmatrix} \; $$
Now set $x_3 = t \in \Bbb R$ arbitrary. From the second row, we get $x_2 = t$, and from the first row
$$ x_1 = 2x_2 - x_3 = 2t - t = t \; ,$$
so we find
$$ W = \{ (t,t,t) \, : \, t \in \Bbb R \} .$$
Now we see, that $ \mathcal B = \{ (1,1,1) \}$ is a basis for $W$, because it's clearly linearly independant and the vector $(1,1,1)$ spans the whole vector space $W$.
To find a basis for a quotient space, you should start with a basis for the space you are quotienting by (i.e. $U$). Then take a basis (or spanning set) for the whole vector space (i.e. $V=\mathbb{R}^4$) and see what vectors stay independent when added to your original basis for $U$.
In your case, it's pretty easy to see that $\{ (2,-1,0,0), (0,0,3,-1) \}$ form a basis for $U$ (essentially $U$ is a null space - you could use Gaussian elimination to find such a basis). We now need $\dim(\mathbb{R}^4)-\dim(U)= 4-2=2$ more vectors to complete this to a basis for $\mathbb{R}^4$. It should be fairly obvious that $(1,0,0,0)$ and $(0,0,1,0)$ are independent of our basis for $U$ (to verify you could always chuck all of these vectors into a matrix, row reduce, and see that you get the identity matrix).
Ok. We now have a basis $\{ (2,-1,0,0), (0,0,3,-1), (1,0,0,0), (0,0,1,0) \}$ for $\mathbb{R}^4$ where the first two vectors form a basis for $U$. The second two vectors are independent of the first two. These vectors give representatives for cosets which form a basis for $\mathbb{R}^4/U$. That is, $\{ (1,0,0,0)+U, (0,0,1,0)+U \}$ is a basis for $\mathbb{R}^4/U$.
Here's more or less an algorithm: Suppose $U$ is a subspace of $V$. Let $A=[a_1\;a_2\;\cdots\;a_m]$ be a matrix whose columns span $U$. Let $B = [b_1\;b_2\;\cdots\;b_n]$ be a matrix whose columns span $V$. Row reduce the matrix $[A\;B] = [a_1\;\cdots\;a_m\;b_1\;\cdots\;b_n]$ to locate pivot columns. Say among the $a_i$'s, $c_1,\dots,c_k$ are pivots and among the $b_j$'s, $d_1,\dots,d_\ell$ are pivots. Then $\{c_1,\dots,c_k\}$ form a basis for $U$ and $\{d_1+U,\dots,d_\ell+U\}$ form a basis for $V/U$.
Best Answer
Rouché-Capelli theorem grants us that if $n$ is the number of equations that characterize the subspace $W$, we have $$\dim(W)=\dim(\mathbb{R^5})-n=4$$
Let v be the vector
$$ v=\ (x_1,x_2,x_3,x_4,x_5)^T $$
If $v \in W$, it must satisfy $x_1-x_3-x_4=0 \implies x_1=x_3+x_4$, so a vector in $W$ must be of the kind
$$(x_3+x_4,x_2,x_3,x_4,x_5)$$
Can you find $4$ linearly independent vectors that satisfy the above constraint?