The error terms need not be unimodal. Let $x$ be a point which is in $A_i$ for exactly $n \geq 1$ indices $i \in [k] := \{1,\ldots,k\}$. Knowing $n$, it's not actually that hard to give a formula for $x$'s contribution to the error term $R_m$ and I think that this makes what's going on pretty clear. Actually I will use $R_m$ for
$ \left| \bigcup_{i=1}^k A_i \right| - \left( \sum_{r=1}^m (-1)^{r+1} \sum_{I \in \binom{[k]}{r}} \left| \bigcap_{i \in I} A_i \right| \right)$
ie. the signed error which results when we count only the contribution to the inclusion-exclusion formula from $r$-fold intersections, $1 \leq r \leq m$.
Since $n \geq 1$, $x$ should contribute exactly $1$ to $| \bigcup_{i =1}^k A_i |$. For positive integers $r$, the number of $r$-element subsets $I$ of $[k]$ for which $x \in \bigcap_{i \in I} A_i$ is $\binom{n}{r}$ (choosing amoung the $n$ indices $i$ for which $x \in A_i$). Thus $x$'s contribution to $R_m$ is simply
$ 1 - \left[ \binom{n}{1} - \binom{n}{2} + \ldots \pm \binom{n}{m} \right] = \sum_{r = 0}^m (-1)^r \binom{n}{r} = (-1)^m \binom{n-1}{m}$
where the closed form can be obtained using the identity $\binom{a+1}{b+1} = \binom{a}{b} + \binom{a}{b+1}$ and induction. Unfortunately I can't see a combinatorial proof. Now it is clear that $x$'s contribution to $R_m$ is most significant when $m \sim (n-1)/2$ and is $0$ for $m \geq n$.
It's easy to disprove unimodality now. For example, if we set things up so that the universe consists of a large number $N$ of points which are in $A_i$ for say $10$ of the $i \in [k]$ and a single further point which is in $A_i$ for $100$ of the $i \in [k]$, then $|R_m|$ will have one a peak around $m=4,5$ when the $N$ points have the most effect. The additional point will contribute the most to the error when $m \sim 50$ and this will result in another peak in $R_m$ since the $N$ points will have long since stopped having any effect after $m=10$.
Added: As Gerry's answer shows, my proposed counterexample above is quite suboptimal. I hadn't played around with the numerics at all when I gave it so I made the numbers big out of some vague paranoia about edge effects.
I agree that Gerry's example simultaneously minimizes the number of sets in the collection and the number of elements in the universe. If $c_n \geq 0$ is the number of elements in exactly $n$ sets of the collection for $n=1,2,\ldots,k$ then, from the argument above, we have
$R_m = (-1)^m \sum_{n=1}^k c_n \binom{n-1}{m}$
So, the possibilities for the sequence of absolute remainders are exactly the nonnegative integer linear combinations of the rows of the array:
$\begin{array}{clcr} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 2 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 3 & 3 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 4 & 6 & 4 & 1 & 0 & 0 & 0 & \cdots \\ 5 & 10 & 10 & 5 & 1 & 0 & 0 & \cdots \\ 6 & 15 & 20 & 15 & 6 & 1 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$
For example, the absolute remainders from Gerry's example are obtained by taking nonzero coefficents 6, 4 and 1 for the 2nd, 3rd and 7th rows. I think that viewed this way it is perhaps easier to convince oneself that this is the "minimal" choice of coefficients such that the resulting linear combination exhibits non unimodality.