As you point out yourself, the Lie bracket is in general not associative. Instead one has the Jacobi identity which tells us that
$$[h_1, [h_2, g]] + [h_2, [g, h_1]] + [g, [h_1, h_2]]].$$
Now, that $h_1, h_2$ commute means by definition that $[h_1, h_2]=0$, hence the last term is $=0$, hence
$$[h_1, [h_2, g]] = - [h_2, [g, h_1]].$$
Now use anticommutativity and bilinearity of the Lie bracket to write this as
$$= -[h_2, -[h_1, g]] = [h_2, [h_1,g]].$$
Or, since you write "adjoint action", maybe you know already (as follows from the Jacobi identity) that the adjoint action defines a Lie algebra representation, i.e.
$$ad([x,y]) =ad(x) \circ ad(y) -ad(y) \circ ad(x) \qquad \text{for all } x,y \in g,$$
and now you apply this to $h_1, h_2$ with $[h_1, h_2] = 0$.In short, if two Lie algebra elements commute, their adjoints commute.
Yes. See Theorem 6.94 of Knapp, Lie groups, beyond an introduction.
This states: if $\mathfrak{g}$ is a real simple Lie algebra then either its complexification $\mathbf{C} \otimes_{\mathbf{R}} \mathfrak{g}$ is simple or its complexification is not simple, $\mathfrak{g}$ is a complex simple Lie algebra, and its complexification is isomorphic to $\mathfrak{g} \oplus \mathfrak{g}$. In either case, $\mathfrak{g}$ is a real form of a semi-simple Lie algebra.
Best Answer
So first things first the "semisimple Lie algebras" tag isn't very big so that might be why you are struggling to find people to answer your question (I came here from the Mathoverflow post but I think this is the right place for this question).
Secondly there are two decompositions often called the Cartan decomposition (you can see this in the comments on your mathoverflow post). One of these is the root space decomposition $\mathfrak{g} = \mathfrak{h}\oplus \bigoplus_{\alpha\in \Delta}\mathfrak{g}^\alpha$ but since this is very intimately linked to the classification via roots and Dynkin diagrams, I believe you must be referring to the other one.
The second one is the decomposition corresponding to a Cartan involution (I think this one is the better one to have the name Cartan decomposition). That is we decompose $\mathfrak{g}$ into a compact and noncompact part:
$$ \mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p} $$
The reason you might not see this at first when you glance through a treatment of semisimple Lie algebras is that they often focus on the complex Lie algebras first. The Cartan decomposition becomes more important when we start discussing real Lie algebras. They are really useful when it comes to classifying things about real Lie algebras.
For example, every Cartan subalgebra of a real Lie algebra is conjugate to one which intersects with the Cartan decomposition. i.e: $$ \mathfrak{h} = (\mathfrak{h}\cap\mathfrak{k}) \oplus (\mathfrak{h}\cap\mathfrak{p}) $$ Then two such $\mathfrak{h}$ are conjugate over $G$ if and only if they are conjugate over $K$ which simplifies understanding conjugacy classes of Cartan subalgebras in $\mathfrak{g}$ to understanding conjugacy classes of the "compact parts" of Cartan subalgebras in $\mathfrak{k}$.