For a prime $p$, the Lie algebra $\mathfrak{su}(p)$ can be decomposed into an orthogonal direct sum of $p+1$ Cartan subalgebras as follows. Consider the clock and shift matrices — these are a pair of order-$p$ matrices which do not commute in $\mathrm{SU}(p)$, and hence are not simultaneously diagonalizable, but do commute in $\mathrm{PSU}(p)$. In fact, they generate a maximal abelian subgroup of $\mathrm{PSU}(p)$, which is elementary of order $p^2$; I'll call this subgroup $T_p \subset \mathrm{PSU}(p)$ to remind you of a Cartan torus. The normalizer of $T_p$ acts transitively on the nonzero elements of $T_p$ (they are all conjugate). Decompose $\mathfrak{su}(p)\otimes \mathbb C$ into eigenspaces for the $T_p$: the weight-zero eigenspace vanishes because $T_p$ is maximal abelian, and the others all have the same dimension because of the transitivity of the normalizer. So each eigenspace is $1$-dimensional. Each line in $T_p^\vee := \hom(T_p, \mathrm{U}(1)) \cong \mathbb{F}_p^2$ sums to a $(p-1)$-dimensional Lie subalgebra of $\mathfrak{su}(p)$ because of the gradings. In fact, this subalgebra is commutative, hence a Cartan. All together, we find $$ \mathfrak{su}(p) = \bigoplus_{x \in \mathbb{F}_pP^1} \text{Cartan}$$ where $\mathbb{F}_pP^1$ is the projective line over $\mathbb{F}_p$. This is an orthogonal direct sum because the Killing form is $T_p$-invariant, and hence only pairs eigenvectors of opposite weights.

I know a similar construction in a few other cases. The exceptional group $G_2$ contains a maximal abelian subgroup $T_2 \subset G_2$ which is elementary of order $2^3$. (For example, the grading group of the octonian algebra $\mathbb{O}$.) Again its normalizer is transitive, and so $\mathfrak{g}_2$ decomposes into a sum of weight spaces indexed by the $7$ nonzero elements of $T_2^\vee$, and all the nonzero weight spaces are the same dimension. Hence this dimension is $2$, and they are all Cartans. Similarly, $F_4$ contains a maximal abelian subgroup which is elementary of order $3^3$, and $E_8$ contains a maximal abelian subgroup which is elementary of order $5^3$. By summing over the lines in these three-dimensional spaces, you find: $$ G_2 = \bigoplus_{x \in \mathbb{F}_2P^2} \text{Cartan}, \qquad F_4 = \bigoplus_{x \in \mathbb{F}_3P^2} \text{Cartan}, \qquad E_8 = \bigoplus_{x \in \mathbb{F}_5P^2} \text{Cartan}.$$ Again these direct sums are orthogonal because the Killing form only pairs opposite weights.

$E_8$ also has another maximal abelian subgroup which is elementary of order $2^5$. This provides an orthogonal direct sum decomposition as $$ E_8 = \bigoplus_{x \in \mathbb{F}_2 P^4} \text{Cartan}.$$ These two direct sum decompositions of $E_8$ into Cartans are not equivalent.

Is this how every orthogonal direct sum decomposition is built? Under some transitivity and/or irreducibility assumptions, it seems from Alexei Kostrikin and Pham Huu Tiep, Orthogonal Decompositions and Integral Lattices, 1994 that the answer is yes. (I do not currently have access to more than the first page of each chapter, so I cannot check the details.) That book also indicates that, as of 1994, the general question of even whether $\mathfrak{su}(6)$ admits an (necessarily non-irreducible) orthogonal direct sum decomposition was open — this is rather whimsically called the "Winnie the Pooh problem", due, apparently, to a very funny Russian pun about $A_5$ that I have not been able to work out.

Has progress been made since 1994?