I share your view that this is a subtle point. To illustrate it, my co-author Dan Seabold and I had pointed to the case of adding a Cohen subset to $\omega_1$ (see example 44 in Boolean ultrapowers paper, the paper you mention). If you use the natural tree order $2^{<\omega_1}$, then this partial order has very few countable maximal antichains, and every ultrafilter, which is to say maximal filter, in the partial order will meet all of them. But the Boolean completion of this order, on the other hand, has many more countable maximal antichains, and no ultrafilter in $V$ will meet them all; furthermore, ultrafilters in the partial order do not generally generate ultrafilters in the Boolean algebra. It was this example that convinced us that the Boolean ultrapower was at bottom about the complete Boolean algebra and not about the underlying partial order.
(Meanwhile, section 8 of the paper tries to recover what one can about the Boolean ultrapower from the partial order. See also the tutorial lecture series I gave for the Young Set Theory workshop in Bonn 2011.)
So, a central issue for your example is the non-completeness of the Boolean algebra
$\newcommand\B{\mathbb{B}}\B$ in the extension. The general
phenomenon, I claim, is that a Boolean algebra $\B$ is never complete after
forcing with it. In truth I was confused about this issue for years, and I am pleased now to have a reasonably clear account of it, in the theorem below. Only a special case of this result is proved in the BU paper.
An important secondary issue is that in many cases, the generic filter $G$ will not even generate an ultrafilter on the completion of $\B$ in the extension.
Theorem. No nontrivial Boolean algebra $\B$ is complete after
forcing with $\B$.
Proof. Suppose $\B$ is a non-atomic Boolean algebra and that
$G\subset V$ is $V$-generic filter. We consider $\B$ in the forcing
extension $V[G]$, and I claim that $\B$ is not complete in $V[G]$.
Some special cases of this are a bit easier to see, and so let us
warm up with these cases before giving the general argument.
For example, consider the case where the forcing adds a new real
$u\subset\omega$. In this case, let $A\subset\B$ be any countably
infinite antichain in $\B$. (Every nontrivial Boolean algebra has a
countably infinite antichain, by a result of Tarski.) Enumerate $A$
as $\{a_n\mid n\in\omega\}$ and by completeness, let
$a=\bigvee_{n\in u} a_n$. Note that $a$ has nonzero overlap with
$a_n$ if and only if $n\in u$, and so we can reconstruct $u$ in $V$
from $a$ and $A$, contradicting the fact that $u$ is new.
An essentially similar idea works if the forcing adds a new subset
$u\subset\kappa$ and $\B$ has an antichain of size $\kappa$ in $V$.
Another common case occurs when the generic filter $G$ is generated
by a linearly ordered set. This happens, for example, whenever one
forces with a tree, for then the generic filter is a branch, which
is linearly ordered. Lemma 43 in the Boolean ultrapower paper shows that a linearly ordered set in a complete
Boolean algebra can never generate an ultrafilter. The argument is
this: $1=b_0>b_1>\cdots>b_\alpha>\cdots$ be a continuous descent
co-initial in the ultrafilter. Let $d_\alpha=b_\alpha-b_{\alpha+1}$
be the corresponding difference antichain, which is maximal. Let
$x=\bigvee\{d_\alpha\mid\alpha\text{ odd}\}$, so $\neg
x=\bigvee\{d_\alpha\mid\alpha\text{ even}\}$. Neither $x$ nor $\neg
x$ is above any particular $b_\alpha$, contradicting the assumption
that they generate an ultrafilter. So if the generic filter $G$ is
generated by a linearly ordered set, as it often is, then the Boolean algebra
cannot be complete in the extension. And worse: the generic filter $G$ does not even generate an ultrafilter on the Boolean completion of $\B$ in the forcing extension $V[G]$!
Finally, I claim that a similar idea allows one to handle the
general case. Assume $G\subset\B$ is $V$-generic for a nontrivial
forcing notion $\B$, and assume $\B$ is complete in the extension
$V[G]$. In the ground model $V$, let us fix an arbitrary nontrivial
splitting assignment of all conditions, $$b=(b)_0\vee (b)_1,$$
with
$(b)_0$ and $(b)_1$ incompatible and nonzero.
The filter $G$ determines a certain descent through these
conditions: we start with $p_0=1$, and then given $p_\alpha\in G$,
it must be that $G$ selects either $(p_\alpha)_0$ or
$(p_\alpha)_1$, which we define as $p_{\alpha+1}$. At limits, we
take the infimum $p_\lambda=\bigwedge_{\alpha<\lambda}p_\alpha$ and
continue, as long as this remains nonzero. This process yields a continuous descent $\vec p=\langle p_\alpha\mid\alpha<\gamma\rangle$ in $G$,
with $\bigwedge_\alpha p_\alpha=0$.
This descent $\vec p$ cannot be in the ground model, since the
difference antichain $d_\alpha=p_\alpha-p_{\alpha+1}$ would be a
maximal antichain missed by $G$.
By the presumed completeness of $\B$ in $V[G]$, let
$$a=\bigvee\left\{(p_\alpha)_0\strut\mid (p_\alpha)_1\in G\right\}.$$
That is, the condition $a$ is the join of the conditions going to
the left, when the generic filter opted to go right. Notice that
this is simply the join of part of the difference antichain, since
$(p_\alpha)_0=p_\alpha-(p_\alpha)_1=d_\alpha$, when
$(p_\alpha)_1\in G$. Because of this, given $p_\alpha\in G$, it
follows that $(p_\alpha)_0\leq a\iff (p_\alpha)_1\in G$. Thus, from
$a$ and the splitting function, we can reconstruct the descent
$\vec p$ in the ground model, contrary to our earlier observation
that it was not in the ground model. So $\B$ is not complete in
$V[G]$. $\Box$.
