You know that $W_1 + W_2 = \operatorname{span} (W_1 \cup W_2)$, so the set $$\{(1,0,1,0),(3,0,1,0), (1,2,3,4),(2,1,1,2)\}$$
certainly spans $W_1 + W_2$.
Also, $$(2,1,1,2) = -\frac32(1,0,1,0) + (3,0,1,0) + \frac12(1,2,3,4)$$
and $\{(1,0,1,0),(3,0,1,0), (1,2,3,4)\}$ is linearly independent so it is a basis for $W_1 + W_2$.
We have:
$$\dim (W_1 \cap W_2) = \dim W_1 + \dim W_2 - \dim (W_1 \cap W_2) = 1$$
so it suffices to find one nontrivial vector in $W_1 \cap W_2$. After inspection, we see that it is:
$$-\frac32(1,0,1,0) + (3,0,1,0) = \left(\frac32, 0, -\frac12, 0\right) = -\frac12 (1, 2, 3, 4) + (2, 1, 1, 2)$$
so $\left\{\left(\frac32, 0, -\frac12, 0\right)\right\}$ is a basis for $W_1 \cap W_2$. You can also take $\left\{\left(3, 0, -1, 0\right)\right\}$ as a basis to get rid of fractions.
Imagine an anologous situation in three dimensions, which is easier to visualize: Let $V$ be a plane through the origin and $W$ a nonzero vector that’s not normal to the plane. The orthogonal complement of $W$ is another plane that intersects $V$ in a line that’s orthogonal to $W$. However, by construction there are elements of $V$ that aren’t orthogonal to $W$, and we can find a pair of them that span $V$.
As to a simpler method of solving this, first verify that the generating vectors of $V$ are linearly independent (they are). You’re looking for elements of $V$ that are orthogonal to $W$, which can be expressed by the equation $$[1,-3,2,1](a[2,2,2,1]^T+b[2,1,1,0]^T+c[5,4,1,1]^T)=0.$$ Expanding and simplifying this equation produces $a+b-4c=0$. This has two free variables, so the intersection is two-dimensional.
Best Answer
Finding the kernel of $M$ is still the right thing to do in this problem. The vectors in the kernel aren't vectors in $U\cap W$ though; as you correctly point out, they are in $\Bbb R^5$ while $U\cap W \subset \Bbb R^4$. Instead, the vectors in the kernel give you the coefficients for a linear combination of the basis vectors which is equal to a vector in $U\cap W$.
By way of example, the kernel of $M$ for the specific $U$ and $W$ you give above is $\textbf{span}\{(-3,-1,4,-1,1)\}$ In other words, $$-3(1,0,1,1) + -1(0,1,0,-1)+4(1,0,1,0)-1(1,0,0,0)+1(0,1,−1,2) = (0,0,0,0)$$ Therefore $$-3(1,0,1,1) + -1(0,1,0,-1)+4(1,0,1,0) = 1(1,0,0,0)-1(0,1,−1,2)$$ and therefore this vector is in $U\cap W$. In fact it spans $U\cap W$ since that intersection happens to be one-dimensional.
More generally, you would find a basis for the kernel of $M$, and for each basis vector, you would come up with a vector of $U\cap W$ in this fashion. The resulting vectors would be a basis of $U\cap W$.