Relating the Lindbladian to Kraus operator: why do we assume the specific Kraus: $K_0=I+L_0 dt$ and $K_{\alpha \neq 0}=\sqrt{dt} L_{\alpha}$ In open quantum systems, for a Markovian evolution, we can derive a Lindblad form for the evolution.
There is a way to relate this Lindblad form to the Kraus decomposition of the quantum map describing the evolution betwee $\rho(0)$ and $\rho(dt)$. Here, I follow the derivations in this lecture note.
By Taylor expansion around $t=0$, we have:
$$\rho(dt)=\rho(0)+\dot{\rho}(0) dt + O(dt^2).$$
We can also apply the Kraus decomposition:
$$\rho(dt)=\sum_{\alpha} K_{\alpha}(dt) \rho(0) K_{\alpha}^{\dagger}(dt)$$
Our goal is now to relate the two expressions. What I would do for this is to consider the general formula:
$$K_{\alpha}(dt)=K_{\alpha}^0+K_{\alpha}^1 \sqrt{dt}+K_{\alpha}^2 dt +...$$
However, in this reference (and in many others in quantum system), they make a choice to consider:
$$K_0=I+L_0 dt$$
$$K_{\alpha \neq 0}=\sqrt{dt} L_{\alpha}.$$
I agree that with this choice we would have a perturbative expansion compatible with the physics, but this is only a specific choice. I could also imagine that all my $K_{\alpha}$ have a non-zero 0'th order term.
Why is this "ad-hoc" consideration for the Kraus operator valid? Is there any book that shows that by doing this "ad-hoc" assumption we are not forgetting any particular scenario that could occur? Is there any simple proof for that?
 A: You are correct that the given form of $K_\alpha$ is not the most general one. However, more general choices lead to the same form of the Lindblad equation.

I will follow the logic of the linked lecture notes: I assume that we already know that a quantum map is described by Kraus operators, and we want to derive the most general possible form of a differential equation describing a quantum map. We therefore want to have
$$ \rho(0) + \dot\rho(0)\, \mathrm dt = \sum_\alpha K_\alpha(\mathrm dt) \rho(0) K_\alpha(\mathrm dt)^\dagger \tag 1 $$
for some Kraus operators $K_\alpha$ to order $\mathrm dt$.
Note that they must satisfy
$$ 1 = \sum_\alpha K_\alpha(\mathrm dt)^\dagger K_\alpha(\mathrm dt) . \tag 2 $$
As you suggest, a reasonable general ansatz for the time dependence of the Kraus operators would be
$$ K_\alpha(\mathrm dt) = A_\alpha + B_\alpha\, \sqrt{\mathrm dt} + C_\alpha\, \mathrm dt . \tag 3 $$
Let us plug (3) into (1) and (2) and compare both sides order for order in $\mathrm dt$.
At zeroth order, we find that $\rho(0) = \sum_\alpha A_\alpha \rho(0) A_\alpha^\dagger$ with $\sum_\alpha A_\alpha^\dagger A_\alpha = 1$.
As shown for example here on qc.se, that is only possible if all $A_\alpha$ are multiples of the identity, $A_\alpha = \lambda_\alpha 1$ with $\sum_\alpha |\lambda_\alpha|^2 = 1$. It is worth plugging this result into our $K_\alpha$ and rewriting (1) and (2) with the new information:
$$ \dot\rho(0)\, \mathrm dt = \Bigl( \rho(0) X + X^\dagger \rho(0) \Bigr)\, \sqrt{\mathrm dt} + \Bigl( \rho(0) Y + Y^\dagger \rho(0) + \sum\nolimits_\alpha B_\alpha \rho(0) B_\alpha^\dagger \Bigr)\, \mathrm dt \tag{1'} , $$
$$ 0 = \Bigl( X + X^\dagger \Bigr)\, \sqrt{\mathrm dt} + \Bigl( Y + Y^\dagger + \sum\nolimits_\alpha B_\alpha^\dagger B_\alpha \Bigr)\, \mathrm dt . \tag{2'} $$
Here, I defined $X = \sum_\alpha \lambda_\alpha B_\alpha^\dagger$ and $Y = \sum_\alpha \lambda_\alpha C_\alpha^\dagger$.
By comparing the terms at order $\sqrt{\mathrm dt}$, you can convince yourself that $X = 0$ must hold*. At order $\mathrm dt$, we find $Y = \mathrm i H - \sum_\alpha B_\alpha^\dagger B_\alpha / 2$ for some Hermitian operator $H$, and we thus obtain the same Lindblad equation:
$$ \dot\rho(0) = -\mathrm i [H, \rho(0)] + \sum_\alpha \Bigl( B_\alpha \rho(0) B_\alpha^\dagger - \frac 1 2 B_\alpha^\dagger B_\alpha \rho(0) - \frac 1 2 \rho(0) B_\alpha^\dagger B_\alpha \Bigr) . $$
*Edit: The most general case would actually be that $X = x 1$ with $x \in i \mathbb R$. That doesn't affect the Lindblad equation, of course.

The form of Kraus operators assumed in the lecture notes is a convenient choice for guaranteeing $X=0$. The derivation there also shows that for any Lindblad equation, one can find Kraus operators (to order $\mathrm dt$) of that form. Any other Kraus representation of the same Lindblad equation must therefore be related by an isometry, as noted in this answer to the question linked above.
As a concrete example, a valid choice of Kraus operators would be
$$ \begin{aligned} K_1 &= \frac{1}{\sqrt 2} + L\, \sqrt{\mathrm dt} - \sqrt 2 L^\dagger L\, \mathrm dt , \\ K_2 &= \frac{1}{\sqrt 2} - L\, \sqrt{\mathrm dt} . \end{aligned} $$
