By definition: If $M$ is a completely reducible $L$-module where $L$ is a Lie algebra, then $M= M_1 \oplus \cdots \oplus M_k$ where each $M_i$ is irreducible $L$-submodule.
Is this direct sum unique?
I know that if $L$ is a semisimple Lie algebra, then $L$ can be written uniquely as in terms of ideals $L=L_1 \oplus \cdots \oplus L_k$ where each $L_i$ is simple (as a Lie-algebra) and each simple ideal of $L$ is one of the $L_i$.
The direct sum is unique if and only if the $M_i$ are pairwise non-isomorphic.
If $M_i \cong M_j$ for some $i \neq j$, then pick a basis $m_{i1}, \ldots, m_{in}$ of $M_i$ and $m_{j1}, \ldots, m_{jn}$ of $M_j$ such that the action of the Lie algebra on $M_i$ and $M_j$ are by the same matrices. Then the vectors $m_{i1} + m_{j1}, \ldots, m_{in} + m_{jn}$ also span a submodule isomorphic to $M_i$, and $m_{i1} - m_{j1}, \ldots, m_{in} - m_{jn}$ do as well, giving a different decomposition of $M_i + M_j$ into irreducibles. There is nothing special about $ (+1, +1)$ and $(+1, -1)$ except that they form an invertible matrix, and (at least if the base field is algebraically closed) the automorphisms of $M_i + M_j$ are precisely $\operatorname{GL}_2$.
It is sometimes easier conceptually to consider the isotypic components of a completely reducible representation $M$. Say that $V_1, V_2, \ldots$ is a complete list of non-isomorphic irreducible modules, then we can define $M(V_i)$ to be the sum of all subrepresentations isomorphic to $V_i$. Then there is a canonical direct sum decomposition $M = M(V_1) \oplus M(V_2) \oplus \cdots$ (you should think of this like an eigenspace decomposition) and each $M(V_i)$ is either zero, isomorphic to $V_i$, or can be non-canonically split up into a direct sum of $V_i$.