Lie bracket of two derivations. Is it composition of derivations or is it multiplication inside the algebra Let $A$ be a $K$-algebra ($K$ a field). Let $\delta:A\to A$ be a $K$-derivation. We say that these $K$-derivations form a Lie algebra by the commutator bracket.
But $[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$, is this by composition or piecewise, or what? It feels very lazy for them to write 'by the commutator bracket' since a priori, the derivations were only defined as a vector space, so talking about the commutator seems ambiguous.
Since $A$ is a $K$-algebra both:
$\delta_1(\delta_2(f))\in A$ and $\delta_1(f)\delta_2(f)\in A$, so I am confused which it is. If the algebra $A$ is commutative, which some people require in the definition of an algebra, then the latter always vanishes, so I would guess that's nonsense, which suggests to me that it is indeed composition. Am I right?

Thanks to Ziggurism above.
$\text{End}_K(A)$ is a ring/associative algebra under composition, and as a result, it makes sense to talk about the commutator. Indeed, the commutator makes use of the multiplication of the algebra, and thus above it is indeed composition, rather than pointwise multiplication.
