Question: $$U = \{(x,y,z,t) \in \mathbb{R}^4 | x + 5y + 4z + t = 0 , y + 2z + t = 0 \} $$

$$W = \{(x,y,z,t) \in \mathbb{R}^4 | x + z + 3t = 0, 2x-3y-4z+3t = 0\} $$

$U \oplus W = \mathbb{R}^4$?

This is my solution so far (which I am not certainly sure that is how I supposed to solve it):

$U = \{(14z + 6t, -t-2z, z, t) \in \mathbb{R}^4 | z,t \in \mathbb{R} \}$

$W = \{(-z-3t, -t-2z, z, t) \in \mathbb{R}^4 | z,t \in \mathbb{R} \}$

Now I know that in order to find $U \oplus W$, first I need to find $U+W$
then to check if $U \cap W = \{0\}$

So I first try to find

$U+W = (14z+6t-z-3t, -t-2z-t-2z, 2z, 2t) = Sp\{(13z + 3t, -2t-4z, 2z, 
2t)\} = \mathbb{R}^4$

is that correct?
also, how do I find $U \cap W$ ?

$U \cap W = ?$

Answer: 
It is enough to see if $U\cap W=0$ or not, i.e. if the system of $4$ equations that define $U\cap W$ has rank $4$ or not? Now, row reduction leads to:
$$\begin{bmatrix}1&5&4&1\\0&1&2&1\\1&0&1&3\\
2&-3&-4&3\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&5&4&1\\0&1&2&1\\0&0&7&7\\0&0&0&0\end{bmatrix}$$
Thus the system of equations has rank $3$,and $\dim(U\cap W)=1$, hence by the rank-nullity theorem, $\dim(U+W)=3$.
You can try finding bases of $U$ and $W$; the equations defining $U$ simplify to
$$
\begin{cases}
x=6z+4t\\
y=-2z-t
\end{cases}
$$
so a basis is $\{(6,-2,1,0),(4,-1,0,1)\}$. (Note that your computation for $U$ is wrong.)

Similarly, a basis for $W$ is $\{(-1,-2,1,0),(-3,-2/3,0,1)\}$.

A spanning set for $U+W$ is thus the union of these two sets. Does the matrix
$$
\begin{bmatrix}
6 & 4 & -1 & -3\\
-2 & -1 & -2 & -2/3\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1
\end{bmatrix}
$$
have rank $4$?

Alternatively, does the system
$$
\begin{cases}
x + 5y + 4z + t = 0\\
y + 2z + t = 0 \\
x + z + 3t = 0\\
 2x-3y-4z+3t = 0
\end{cases}
$$
have non trivial solution?