We have the vector subspaces of $\mathbb{R}^4$ $$A=\{(x,y,2z,z) \mid x, y, z\in \mathbb{R}\}=\{x(1,0,0,0)+y(0,1,0,0)+z(0,0,2,1)\mid x, y, z\in \mathbb{R}\}$$ $$B=\{(x,y,z,y)\mid x,y,z\in \mathbb{R}\}=\{x(1,0,0,0)+y(0,1,0,1)+z(0,0,1,0)\mid x,y,z\in \mathbb{R}\}$$
A basis of $A$ is therefore, $\{(1,0,0,0), (0,1,0,0), (0,0,2,1)\}$, so it is of dimension $3$.
A basis of $B$ is therefore, $\{(1,0,0,0), (0,1,0,1), (0,0,1,0)\}$, so it is of dimension $3$.
Right?
How can we find a basis and the dimension of $A\cap B$ ?
Is the sum $A+B$ all the sums $a+b$ where $a\in A$ and $b\in B$ ? And how can we find a basis and the dimension?
A vector $u = (x, y, z, t) \in A \cap B$ iff $z=2t$ and $y=t$. This makes a system of 2 equations in the 4 unknown $x, y, z, t$. This system has rank 2 and one can write $u = (x, t, 2t, t)$ where $x, t$ are arbitrary parameters. Thus $u = x (1,0,0,0) + t (0,1,2,1)$. This gives the required basis.
$A+B$ is indeed the sum of all $a+b$ where $a\in A$ and $b\in B$. It is also the set of all linear combinations $x a + y b$ where $x, y \in \mathbb{R}$ and $a\in A$ and $b \in B$. A basis of $A + B$ is very easy to find in your example because $(1, 0, 0, 0)\in A \subset (A+B)$, the same for $(0,1,0,0)$. Similarly, $(0,0,1,0)\in B$ and one has
$$(0,0,0,1) = (0,0,2,1) - 2 (0,0,1,0)\in A+B$$
Hence the whole usual basis of $\mathbb{R}^4$ belongs to $A+B$, so $A+B = \mathbb{R}^4$