(Just so this question has an answer.)

If $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^l(M)$, $\alpha\wedge\beta = (-1)^{kl}\beta\wedge\alpha$.

So if one of $\alpha$ or $\beta$ has even degree (i.e. $k$ or $l$ is even), $\alpha\wedge\beta = \beta\wedge\alpha$; if both $\alpha$ and $\beta$ have odd degree, then $\alpha\wedge\beta = -\beta\wedge\alpha$.

In particular, if $\alpha$ has odd degree, $\alpha\wedge\alpha = -\alpha\wedge\alpha$ so $\alpha\wedge\alpha = 0$. Note, if $\alpha$ has even degree, the above discussion gives the tautology $\alpha\wedge\alpha = \alpha\wedge\alpha$.

In the case of the Lefschetz operator, we have $L\circ L : \Omega^k(M) \to \Omega^{k+4}(M)$ given by $$(L\circ L)(\alpha) = L(K\wedge\alpha) = K\wedge(K\wedge\alpha) = (K\wedge K)\wedge\alpha$$ where $K$ is the fundamental form. As $K$ is a two-form, it has even degree, so we can't say anything about $K\wedge K$; in particular, we cannot deduce that it is zero.

(Just so this question has an answer.)

If $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^l(M)$, $\alpha\wedge\beta = (-1)^{kl}\beta\wedge\alpha$.

So if one of $\alpha$ or $\beta$ has even degree (i.e. $k$ or $l$ is even), $\alpha\wedge\beta = \beta\wedge\alpha$; if both $\alpha$ and $\beta$ have odd degree, then $\alpha\wedge\beta = -\beta\wedge\alpha$.

In particular, if $\alpha$ has odd degree, $\alpha\wedge\alpha = -\alpha\wedge\alpha$ so $\alpha\wedge\alpha = 0$. Note, if $\alpha$ has even degree, the above discussion gives the tautology $\alpha\wedge\alpha = \alpha\wedge\alpha$.

In the case of the Lefschetz operator, we have $L\circ L : \Omega^k(M) \to \Omega^{k+4}(M)$ given by $$(L\circ L)(\alpha) = L(K\wedge\alpha) = K\wedge(K\wedge\alpha) = (K\wedge K)\wedge\alpha$$ where $K$ is the fundamental form. As $K$ is a two-form, it has even degree, so we cannot deduce that it is zero. However, as Deane Yang points out below, we know that $K\wedge K$ is nowhere zero as $K^n$ is a volume form.