Let me try to provide a helpful elementary answer.

Suppose that $F$ is a filter on a set $X$ arising as the intersection of finitely many ultrafilters $$F=\mu_1\cap\cdots\cap\mu_n.$$ We may assume that the $\mu_i$ are distinct. Since the filters $\mu_i$ are different, they must disagree on some subsets of $X$. For each pair $i\neq j$ we can find a set in $\mu_i$ and not in $\mu_j$, and by intersecting these sets, we can produce a partition $$X=X_1\sqcup X_2\sqcup\cdots\sqcup X_n$$ such that each $\mu_i$ concentrates on the $i$th piece $X_i$.

Note that a subset $Y\subseteq X$ is in $F$ just in case $Y\cap X_i\in\mu_i$ for every $i$.

Let me define that that a filter $F$ localizes to an ultrafilter on $Y\subseteq X$ if $\{Z\subseteq Y\mid Z\cup(X\setminus Y)\in F\}$ is an ultrafilter on $Y$.

What we've observed is that a filter $F$ is a finite intersection of ultrafilters if and only if there is a finite partition such that $F$ localizes to an ultrafilter on each piece.

A further conclusion is that if a filter $F$ is a finite intersection of ultrafilters $F=\mu_1\cap\cdots\cap\mu_n$, then the number $n$ is determined from $F$, and furthermore the ultrafilters $\mu_i$ are also determined from $F$.

To see this, if we partition $X$ into more than $n$ pieces, then some piece must not be in any $\mu_i$ and so will have measure $0$ with respect to $F$. So $n$ is determined.

For the measures $\mu_i$ being determined, I claim that $F\subseteq\nu$ for an ultrafilter $\nu$ if and only if $\nu=\mu_i$ for some $i$. If $F$ is contained in $\nu$ and $\nu\neq\mu_i$ for any $i$, then we can find a set $Y\in\nu$, $Y\notin\mu_i$ any $i$. But in this case the complement of $Y$ is in $F$, contrary to $F\subseteq\nu$.

So the ultrafilters that are used are exactly the ultrafilter completions of $F$. Putting this all together, what we have is:

Theorem. The following are equivalent for any filter $F$:

1. $F$ arises as a finite intersection of ultrafilters.
2. $F$ has only finitely many completions to an ultrafilter.
3. There is a finite partition of the underlying set such that $F$ localizes to an ultrafilter on each piece.
4. For some finite $n$, there is a partition of $F$ into $n$ pieces, none $F$-measure 0, but no such partition with $n+1$ pieces.

To see that 4 suffices for the others, if we have $X=X_1\sqcup\cdots X_n$ and each $X_i$ is not $F$-measure 0, then let $\mu_i$ be the localization of $F$ to $X_i$. If this is not an ultrafilter, then we can further partition $X_i$ into two pieces not of measure $0$, and make such a partition of size $n+1$, contrary to 4.

Let me try to provide a helpful elementary answer.

Suppose that $F$ is a filter on a set $X$ arising as the intersection of finitely many ultrafilters $$F=\mu_1\cap\cdots\cap\mu_n.$$ We may assume that the $\mu_i$ are distinct. Since the filters $\mu_i$ are different, they must disagree on some subsets of $X$. For each pair $i\neq j$ we can find a set in $\mu_i$ and not in $\mu_j$, and by intersecting these sets, we can produce a partition $$X=X_1\sqcup X_2\sqcup\cdots\sqcup X_n$$ such that each $\mu_i$ concentrates on the $i$th piece $X_i$ (and therefore gives measure zero to the other pieces).

Note that a subset $Y\subseteq X$ is in $F$ just in case $Y\cap X_i\in\mu_i$ for every $i$.

Let me define that that a filter $F$ localizes to an ultrafilter on $Y\subseteq X$ if $\{Z\subseteq Y\mid Z\cup(X\setminus Y)\in F\}$ is an ultrafilter on $Y$.

What we've observed is that a filter $F$ is a finite intersection of ultrafilters if and only if there is a finite partition such that $F$ localizes to an ultrafilter on each piece.

A further conclusion is that if a filter $F$ is a finite intersection of ultrafilters $F=\mu_1\cap\cdots\cap\mu_n$, then the number $n$ is determined from $F$, and furthermore the ultrafilters $\mu_i$ are also determined from $F$.

To see this, if we partition $X$ into more than $n$ pieces, then some piece must not be in any $\mu_i$ and so will have measure $0$ with respect to $F$. So $n$ is determined.

For the measures $\mu_i$ being determined, I claim that $F\subseteq\nu$ for an ultrafilter $\nu$ if and only if $\nu=\mu_i$ for some $i$. If $F$ is contained in $\nu$ and $\nu\neq\mu_i$ for any $i$, then we can find a set $Y\in\nu$, $Y\notin\mu_i$ any $i$. But in this case the complement of $Y$ is in $F$, contrary to $F\subseteq\nu$.

So the ultrafilters that are used are exactly the ultrafilter completions of $F$. Putting this all together, what we have is:

Theorem. The following are equivalent for any filter $F$:

1. $F$ arises as a finite intersection of ultrafilters.
2. $F$ has only finitely many completions to an ultrafilter.
3. There is a finite partition of the underlying set such that $F$ localizes to an ultrafilter on each piece.
4. For some finite $n$, there is a partition of $F$ into $n$ pieces, none $F$-measure 0, but no such partition with $n+1$ pieces.

To see that 4 suffices for the others, if we have $X=X_1\sqcup\cdots X_n$ and each $X_i$ is not $F$-measure 0, then let $\mu_i$ be the localization of $F$ to $X_i$. If this is not an ultrafilter, then we can further partition $X_i$ into two pieces not of measure $0$, and make such a partition of size $n+1$, contrary to 4.