From what I can tell, most of your question is adequately addressed in this other post, except perhaps for the last part:

The reason I ask this question is that if for any $\lambda \in K^*$, we have $\lambda \ge_{K^*} 0$, why not just using $\lambda \in K^*$ instead of $\lambda \ge_{K^*} 0$ in the above statement?

I can think of two good reasons to prefer the inequality notation here.

First, for *aesthetic consistency*. Which of these two stacks of constraints seems cleaner/easier to read (left or right)?
\begin{array}{ccc}
x \succeq_K y &\qquad& x\succeq_K y \\
y \succeq_K z && y\succeq_K z \\
z \succeq_K 0 && z\in K
\end{array}
I claim the left-hand side is cleaner. I see no reason to drop back to set notation $z\in K$ just in that last case just because the right-hand side happens to be zero.

Second, for *conceptual clarity*. The inequality notation $\lambda \succeq_{K^*} 0$ reminds the reader of the tie to traditional inequalities. Consider this linear program:
\begin{array}{ll}
\text{minimize} & c^T x \\
\text{subject to} & A x \leq b
\end{array}
The Lagrangian for this problem is
$$L(x,z) = c^T x - \langle z, b - Ax \rangle \qquad z\geq 0.$$
Now let's replace the inequality with a generalized inequality:
\begin{array}{ll}
\text{minimize} & c^T x \\
\text{subject to} & A x \preceq_K b
\end{array}
where $K$ is a proper cone. The new Lagrangian is
$$L(x,z) = c^T x - \langle z, b - Ax \rangle \qquad z \succeq_{K^*} 0.$$
where $K^*$ is the dual cone.
Yes, I *could* have written $b-Ax\in K$ and $z\in K^*$ instead; but doing so breaks the conceptual parallels to linear programming. Why do that?