You can find a basis for $U_1$ simply by solving the linear system
$$
\begin{align}
3x_1 + x_2 + 2x_3 - x_4&=0\\
2x_1 + 4x_2 + x_3 - x_4&=0
\end{align}$$
Of course, there are many ways to solve this system, e.g. we can try to use elementary row operations:
$$
\begin{pmatrix}
3 & 1 & 2 & -1 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
1 & 7 & 0 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
-1 & 7 & 0 & 1 \\
\end{pmatrix}
$$
Now we see that the original system is equivalent to the the system of equations $x_1-3x_2+x_3=0$, $-x_1+7x_2+x_4=0$; i.e.
$x_3=-x_1+3x_2$, and $x_4=x_1-7x_2$. We see that for any choice of $x_1$, $x_2$ we obtain a solution of the form
$$(x_1,x_2,-x_1+3x_2,x_1-7x_2)=x_1(1,0,-1,1)+x_2(0,1,3,7),$$
which means that $U_1=[(1,0,-1,1),(0,1,3,7)]$. I.e., these two vectors are the basis for $U_1$. (Of course, there are many different basis for the same subspace. So might get different vectors. It is also useful to check whether the vectors in basis fulfill the original system - if not, there must be a mistake somewhere.)
The space $U_2$ is generated by vectors $L_2(1,0)$ and $L_2(0,1)$, hence $U_2=[(1,1,2,3),(-1,-3,-8,-27)]$. These vectors are linearly independent, so they form a basis. (To check whether two vectors are independent you only need to check whether one of them is a multiple of the other one.)
Now let us try $U_1+U_2$. We know that $U_1+U_2=[(1,0,-1,1),(0,1,3,7),(1,1,2,3),(-1,-3,-8,-27)]$, but we don't know whether these vectors are independent. So we try to make row echelon form.
$$
\begin{pmatrix}
1 & 0 & -1 & 1\\
0 & 1 & 3 & 7\\
1 & 1 & 2 & 3\\
-1 &-3 &-8 &-27
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 1\\
0 & 1 & 3 & 7\\
0 & 0 & 0 & 1\\
-1 &-3 &-8 &-27
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 3 & 0\\
0 & 0 & 0 & 1\\
-1 &-3 &-8 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 3 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{pmatrix}.$$
Now we know that vectors $(1,0,-1,0)$, $(0,1,3,0)$, $(0,0,0,1)$ form a basis of $U_1+U_2$ and that $d(U_1+U_2)=3$.
Since $d(U_1)+d(U_2)=d(U_1+U_2)+d(U_1\cap U_2)$, we know that $d(U_1\cap U_2)=1$.
If we notice that $(2,4,10,30)=(1,1,2,3)+(1,3,8,27)$ and $2(1,0,-1,1)+4(0,1,3,7)=(2,4,10,30)$, we see that the vector $(1,2,5,10)$ belongs to $U_1\cap U_2$. Since $U_1\cap U_2$ is one-dimensional, this implies that $U_1\cap U_2=[(1,2,5,10)]$.
Now let's try to think how we could find the basis for $U_1\cap U_2$ without guesswork.
One possibility is to have a closer look at what we have done when doing row operations. If we denote the rows of the above matrix by $\vec a_1$, $\vec a_2$, $\vec b_1$, $\vec b_2$, then we have can find out (using row operation similar to the above) that:
$\vec a_1+\vec a_2-\vec b_1=(0,0,0,5)$
$\vec a_1+3\vec a_2+\vec b_2=(0,0,0,-5)$
Obviously by combining the above two vectors we get $2\vec a_1+4\vec a_2+\vec b_2-\vec b_1=\vec 0$, which is the same as
$$2\vec a_1+4\vec a_2 = \vec b_1-\vec b_2,$$
hence this vector belongs to the intersection.
The above was still based on the lucky fact that we only had to find one vector. Can we approach this in a more systematic way?
Of course we can. To find vectors in $U_1\cap U_2$ we can solve the linear system of equations given by $x_1\vec a_1+x_2 \vec a_2=y_1\vec b_1+y_2 \vec b_2$. Once we know all possible pairs $(x_1,x_2)$, we can obtain all vectors belonging to $U_1\cap U_2$.
Based on your comment I've guess you got the following when solving the system:
$$
\begin{pmatrix}
3 & 1 & 2 & -1 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 7/10 & -3/10 \\
0 & 1 & -1/10 & -1/10
\end{pmatrix}
$$
From the last matrix you can get that the solutions of this system form the subspace
$U_1=[(-7/10,1/10,1,0),(3/10,1/10,0,1)]=[(-7,1,10,0),(3,1,0,10)]$.
This is another basis for the same subspace $U_1$. (And this is also correct.)
