Convolution of discrete uniform distributions

If $X$ and $Y$ are independent integer-valued random variables uniformly distributed on $[0,m]$ and $[0,n]$ respectively, then the probability mass function (pmf) of $Z = X+Y$ has a trapezoidal shape as you have already noted, and Khashaa has written down for you. The answer can be summarized as follows, but whether this is more compact or appealing is perhaps a matter of taste.

$$P\{Z=k\} = \begin{cases}\displaystyle \frac{k+1}{(m+1)(n+1)},& k \in [0, \min(m,n)-1],\\ \\ \displaystyle\frac{1}{\max(m,n)+1},& k \in [\min(m,n), \max(m,n)],\\ \\ \displaystyle\frac{(m+n)-(k-1)}{(m+1)(n+1)}, & k \in [\max(m,n)+1, m+n].\end{cases}$$

To my mind, the easiest way of solving this problem, and indeed a way that works for dependent and non-uniformly distributed random variables as well, is to write down the joint pmf of $(X,Y)$ as a rectangular array or matrix of $m$ columns (numbered $0, 1, \ldots , m$ from left to right) and $n$ rows (numbered $n, n-1, \ldots, 0$ from top to bottom. Then, $P\{X+Y=k\}$ is the sum of the entries on the $k$-th diagonal of this array. For the case of constant entries, we get the nice trapezoidal shape that the OP has noticed.