In Gauss-Seidel, you are using this $A=L+U$ splitting implicitly. So, you never form $L$, $U$, or $L^{-1}$ explicitly. Which is extremely good, since forming an additional matrix (not even talking about a calculation of an explicit inverse) is a huge burden.
Instead, since $L$ is lower triangular, you can change the explicit inverse of $L$, by performing a row-by-row forward subsitution.
Notice, that solving for (via the forward substitution when $L$ is lower-triangular):
$$
\begin{equation}
L\mathbf{y}=\mathbf{b}
\label{1}
\tag{1}
\end{equation}
$$
is theoretically the same as performing a matrix-vector product:
$$
\begin{equation}
\mathbf{y}=L^{-1}\mathbf{b}
\label{2}
\tag{2}
\end{equation}
$$
However, numerically you always want $\eqref{1}$, not $\eqref{2}$. The reasons are simple: computation of the matrix inverse is numerically unstable and has a huge cost. (see Q1 (especially this answer), Q2 for some additional details).
With that in mind, take a look at the expression from Wikipedia article on Gauss-Seidel. The iteration:
$$
\mathbf{x}^{(k+1)}=L^{-1}(\mathbf{b}-U\mathbf{x}^{(k)})
$$
is totally equivalent to a for-loop for $i$
$$
\begin{equation}
x^{(k+1)}_i=\frac{1}{a_{ii}}\left(b_i-\sum\limits_{j=1}^{i-1}a_{ij}x_{j}^{(k+1)}\right)-\frac{1}{a_{ii}}\left(\sum\limits_{j=i+1}^{n}a_{ij}x_j^{(k)}\right)
\label{3}
\tag{3}
\end{equation}
$$
Here, $a_{ij}$ are the entries of the original matrix $A=L+U$, $k$ is the iteration number, $n$ is the dimension of the square matrix $A$. By looking at $\eqref{3}$, you still see those $L$ and $U$; however, they are present only implicitly, and all operations are done with the entries of the original matrix $A$ stored exactly as it was before.
side-note:
Matrix decomposition usually implies factorization, so $A=L+U$ is not really (strictly) one according to the definition.
$A=L+U$ splitting (used in this question) has nothing to do with a more famous $A=LU$ LU decomposition.
