On dual cones: $(K\cap L)^+\subseteq K^+ + L^+$

It is straightforward that $P^+\cap Q^+ \subset (P+Q)^+$. Conversely if $x \in (P+Q)^+$, assuming that $P$ and $Q$ are non-empty cones you have that $0\in Cl(P)$ and $0 \in Cl(Q)$ whence $x\in P^+$ and $x\in Q^+$.

So we have $P^+\cap Q^+=(P+Q)^+$ for any two non-empty cones $P, Q$.

Taking the dual cone we get that:

$$Cl(P+Q) = (P^+ \cap Q^+)^+$$ so if you let $P=K^+$ and $Q=L^+$ (which are two closed clones) and using the fact that the bidual is the closure you get that: $$K^+ + L^+ = (Cl(K) \cap Cl(L))^+ $$

So the question you are left with is: can the dual cone of $Cl(K)\cap Cl(L)$ be smaller than that of $K\cap L$?

It can be the case for instance if $K$ is a half-line on the boundary of $L$ (open), in which case $K\cap L=\emptyset$ and $Cl(K)\cap Cl(L)=K$.

A sufficient condition is obviously that $K$ and $L$ are closed.

I don't think the condition $K\cap Int(L) \not= \emptyset$ is necessary, for instance you could have $K$ and $L$ two distinct half-lines - they are closed but their interior is empty.

A necessary condition is that $Cl(K)\cap Cl(L)=Cl(K\cap L)$, using again the fact that the bidual is the closure. Because the dual of $K\cap L$ and that of its closure are the same, this condition is also sufficient. Not sure we can make this condition more explicit.