In Humphreys, given a finite dimensional semisimple Lie algebra $L$ and a maximal toral subalgebra $H$, $$L_\alpha := \{x\in L|[hx] = \alpha(h)x\;\forall h\in H\}$$ Then since $ad_L\;H$ is a commuting family of semisimple endormorphisms of $L$ thus can be simultaneously diagonalized. Thus $$L = \bigoplus_{\alpha \in H^*} L_\alpha$$where only finitely many summands are nonzero.
Later we see in fact $L_0 = C_L(H) = H$, and let $\Phi$ be the set of $\alpha$ where $LK_\alpha\ne0$, then $$L = H\bigoplus_{\alpha\in \Phi} L_\alpha$$
So far so good, but I'm confused about a particular set of $\{f_i\}\subset H^*$ that I think will do the job. Fix a basis $\{e_1,\cdots,e_n\}$ of $L$ where $ad_L\;H$ are simultaneously diagonalized. Define $f_i\in H^*$ to be the function that takes $h\in H$ and returns the $i$th diagonal entry of the matrix form of $ad\;h$. Then I think $L = \bigoplus_{i =1}^n L_{f_i}$ will be enough because each $L_{f_i}$ contains the subspace spanned by the $i$th basis vector. (It's possible that some of the $f_i$ concide, then we only count it in the sum once).
Why is this not good? Also, I know $H\subset L_0$, but can't see how $L_0$ sits in $L$ directly in this decomposition $L = \bigoplus_{i =1}^n L_{f_i}$.
You do have that decomposition, because you just said ``Let's take a basis of eigenvectors and for each eigenvector consider the eigenspace that contains it''. The problem is that this does not give you any explicit information about the algebra itself. You can see $L_0$ inside, if you know what $e_i$ lie in $H$. Just look at $L_{f_i}$ in that case.