Let $\mathfrak g$ be a finite dimensional semisimple lie algebra over $\mathbb C$. Let $\{e,h,f\}$ be an $\mathfrak{sl}_2$-triple in $\mathfrak g$ (i. e. with relations $[h,e] = 2e$, $[h,f]=-2f$ and $[e,f]=h$). Why do we have
$$\mathfrak g = [e,\mathfrak g]\oplus {\rm Ker}({\rm ad}f)?$$
Maybe you only need $\mathfrak g$ to be a finite dimensional representation of $\mathfrak {sl}_2$.
On
We have that $\ker ad(f))=Z_{\mathfrak{g}}(f)$ is the centralizer of $f$ in $\mathfrak{g}$, with the direct sum decompositions \begin{align*} \mathfrak{g} & = Z_{\mathfrak{g}}(e)\oplus [f,\mathfrak{g}] \\ & = Z_{\mathfrak{g}}(f)\oplus [e,\mathfrak{g}], \end{align*} which give $\mathfrak{g}=[e,\mathfrak{g}]+\ker ad(f)=[f,\mathfrak{g}]+\ker ad(e)$. A proof follows from $\mathfrak{sl}(2)$-theory, and is not bounded to the adjoint representation. It can be found, for example, in the paper The index of a Lie algebra, the centraliser of a nilpotent element, and the normaliser of the centraliser by Panyushev, Proposition $2.1 (iii)$.
As indicated in the question and the other answer, this is really a fact about $\mathfrak{sl}_2$ representations and doesn't depend at all upon the adjoint representation, so perhaps it's worth observing that it follows from results found in standard references:
First, for a finite dimensional irreducible $\mathfrak{sl}_2$-module $V$ with highest weight vector $v$, we have the direct sum decomposition $$\mathbf{C} v \oplus \mathbf{C}\{fv,f^2v,\dots \},$$ and using the standard relations between $e,f,$ and $h$ it follows that $$\mathbf{C} v=\mathrm{ker}(e) \quad \text{and} \quad \mathbf{C}\{fv,f^2v,\dots \}=\mathrm{im}(f).$$ Now an arbitrary finite dimensional $\mathrm{sl}_2$-module $V$ is a direct sum of irreducibles, so it follows that $$V=\mathrm{ker}(e) \oplus \mathrm{im}(f)$$ for for all finite dimensional $\mathfrak{sl}_2$-modules.
The facts you need about $\mathrm{sl}_2$-representations for this argument to proceed may be found in many places, e.g. Humphrey's book, or the text of Fulton and Harris.