When the inputs are the indicator functions, set non-disjointness is in NPcc with witnesses
from {0,1,2,3,...,n-2,n-1} $\big(\hspace{-0.04 in}$so in particular is in NotTooManyPcc$\hspace{-0.03 in}\big)$, but by this paper,
the coUSBP (and thus also P) communication complexity of that problem is $\Theta$(n).
By padding, that means we leave coUSBPcc $\big(\hspace{-0.04 in}$and thus also Pcc$\hspace{-0.03 in}\big)$ as soon
as the number of witnesses is no longer polylogarithmically-bounded.
Also, for all functions f : {0,1,2,3,...} $\to$ (0,1] , if the version with at most
nf(n) correct witnesses equals NPcc then f has a positive lower bound.
Proof:
Let f be any function from {0,1,2,3,...} to (0,1] such that the problem at
the start of this answer can be solved with at most nf(n) correct witnesses.
Let g : {0,1,2,3,...} $\to$ {0,1,2,3,...} be given by
g(0) = 0 = g(1) and otherwise g(n) = (log(n))1/(f(n)) .
g(0) = 0 and g(1) = 0 < 1 and for all other inputs n,
g(n) = (log(n))1/(f(n)) < n1/(f(n)) ≤ n1/1 = n1 = n .
Accordingly, consider the problem non-disjointness of subsets of {0,1,2,3,...,(g(n))-2,(g(n))-1} whose inputs are padded indicator functions.
By removing the padding, that problem can be solved with at most (g(n))f(n) correct witnesses.
For all integers n such that 1 < n , (g(n))f(n) = $\big(\hspace{-0.05 in}$(log(n))1/(f(n))$\hspace{-0.03 in}\big)^{\hspace{.02 in}f\hspace{.02 in}(n)}$ = (log(n))(1/(f(n)))$\cdot$f(n) = (log(n))1 .
Thus the problem is in FewPcc, so by the result you mentioned, it's also in Pcc.
By the communication lower bound in this answer's starting sentence, that means g is
at most polylogarithmic, so 1/(f(n)) is bounded above, so f has a positive lower bound.
I'm not aware of any other results about limited-ambiguity versions of NPcc,
although with inputs as in the rest of this answer, (paddings of, if necessary) the problems
" Does the intersection have at least _ elements? "
seem like natural candidates for separating those classes.
$\big(\hspace{-0.04 in}$On the other hand, I wasn't even aware of unambiguousPcc = Pcc ,
and still know neither a proof of that nor a reference for that.$\hspace{-0.03 in}\big)$