There is a counter-example againt my conjecture.

I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.

Take $\mathcal{P} \in  F$ from <a href="http://mathoverflow.net/a/190653/4086">the previous counter-example</a>. We have
$$ \forall a \in \prod_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}_i : a
   \notin \mathcal{P} . $$
Take $k = 1$.

Let $\mathcal{L} = \mathcal{P} |_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a
\notin \mathrel{\upuparrows F}$ and thus $a_k \asymp \langle
\upuparrows F \rangle_k a|_{(\operatorname{dom} F) \setminus \{ k \}}$.

Consequently $\mathcal{P}_k \asymp \langle \upuparrows F \rangle_k
a|_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}_k \asymp
\bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F) \setminus \{ k \}}
\operatorname{atoms} \mathcal{L_{}}_i} \langle \upuparrows F \rangle_k a$
because $\mathcal{P}_k$ is principal.

But $\mathcal{P}_k \not\asymp \langle \upuparrows F \rangle_k 
\mathcal{L}$. Thus follows $\langle \upuparrows F \rangle_k 
\mathcal{L} \neq \bigvee^{\mathfrak{F}}_{a \in \prod_{i \in (\operatorname{dom} F)
\setminus \{ k \}} \operatorname{atoms} \mathcal{L}_i} \langle \upuparrows
F \rangle_k a$.