MOD-N Circular Convolution
Write $h(z) = -1 + 3z -2z^2 + z^3$ and compute $h(z) \bmod (z^2 - 1)$, that is,
divide $h(z)$ by $z^2 - 1$ and take just the remainder.  While this looks
very complicated, if you think about it a bit you will see that all you
are doing is dividing $[-1\quad3\ -2\quad 1]$ into $[-1\quad 3]$ and 
$[-2\quad 1]$ and
adding the shorter vectors to get $[-3\quad 4]$.  Repeat for
$x = [1 \ -1\ -2\quad 1 \quad 3 \quad 2\quad 1 \quad 2]$ to add 
four vectors of length $2$ to get 
$[3\quad 4]$.  This is presumably not too hard to do in MATLAB though I am 
not familiar enough with the syntax to suggest
specific commands.  Next, compute the cyclic convolution
of $[-3\quad 4]$ and $[3\quad 4]$ preferably without invoking MATLAB functions.
The result is 
$$[\{(-3)\times 3 + 4\times 4\}\quad \{(-3)\times 4 + 4 \times 3\}]
= [7\quad 0]$$ 

Mathematically, what you are doing is computing $h(z)x(z) \bmod (z^2-1)$
which can be done fancily by first finding $h(z)x(z)$ using FFTs and what 
have you followed by the $\bmod (z^2-1)$ computation (this effectively
chops up the long vector into short pieces and adds them), or more simply
by first computing $\hat{h}(z) = h(z) \bmod (z^2-1)$ and 
$\hat{x}(z) = x(z) \bmod (z^2-1)$ (the chopping up into shorter vectors
and adding them) and then computing the cyclic convolution 
$\hat{z}(z)\hat{x}(z) \bmod (z^2 - 1)$ which is easy to do.


  Chop-add-convolve is easier than convolve-chop-add