Difference of uniform random variables

Let $W=-Y$, then the density of $Z=X+W$ is given by the convolution: \begin{align} f_Z(z) &= (f_X\star f_W)(z)\\ &= \int_{\mathbb R} f_X(x)f_W(z-x)\ \mathsf dx\\ &= \int_{\mathbb R} \mathsf 1_{(0,1)}(x)\cdot\mathsf 1_{(-1,0)}(z-x)\ \mathsf dx\\ &= \begin{cases} \int_0^{1+z}\ \mathsf dx = 1+z,& -1<z<0\\ \int_z^1\ \mathsf dx = 1-z,& 0<z<1. \end{cases} \end{align}

A slight variant on @Math1000's strategy is to use $1-Y\sim U(0,1)$ so $X-Y+1$ has an $n=2$ Irwin–Hall distribution. Another approach is to note $X$ has MGF $\frac{e^t-1}{t}$, so $X-Y$ has MGF $$\frac{e^t-1}{t}\frac{e^{-t}-1}{-t}=\frac{e^t-1-t}{t^2}+t\leftrightarrow -t.$$But $\frac{e^t-1-t}{t^2}$ is the MGF of the distribution with pdf $2(1-x)$ on $[0,\,1]$, since $$\int_0^12(1-x)e^{tx}dx=\frac{1}{t^2}[(2t+2-2tx)e^{tx}]_0^1=\frac{2(e^t-1-t)}{t^2}.$$Therefore, $X-Y$ has pdf $1-|x|$ on $[-1,\,1]$.