I derived this distribution, but I'm not entirely sure of the middle point. Intuitively it should be $\frac{a+b+c+d}{2}$.
For two independent Uniform rvs, $X \sim R[a,b], Y \sim R[c,d]$ derive the distribution of $Z=X+Y$. Using the convolution formula, $$ f(z) = \int_{\mathbb{R}}f_{X}(x)f_{Y}(z-x)dx $$ Since $f$ is $0$ outside of respective intervals, I need to find the interval on which $f_{X}f_{Y}>0$. Therefore, $f_{Z}(z) = 0 \ \forall \ z<a+c$ and $z>b+d$. So, I consider the interval on which $y$ is defined: $$ c<z-x<d $$ Since we integrate wrt to $x$, we need to find the bounds of integration for x. Clearly lb is $a$, and from the inequality above the ub is $x<z-c$: $$ f_1(z) = \int_{a}^{z-c}f_X(x)f_Y(z-x)dx = \frac{z-c-a}{(b-a)(d-c)} $$ For the second interval, the ub is obviously $b$, and the lower bound is $z-d$: $$ f_2(z) = \int_{z-d}^{b}f_X(x)f_{Y}(z-x)dx = \frac{b+d-z}{(b-a)(d-c)} $$ Putting them together, $$ f_{Z}(z) = \left\{ \begin{array}{lr} \frac{z-c-a}{(b-a)(d-c)}&a+c<z<??\\ \frac{b+d-z}{(b-a)(d-c)}&??<z<b+d \\ 0& \text{otherwise} \end{array}\right. $$