Let $X,Y$ be uniformly distributed on $[a,b]$. Then the pdf of $Z = X-Y$ is
$$p(z) = \int_\mathbb{R} dx\int_\mathbb{R} dy \ p(x) p(y) \ \delta(z-(x-y)) =$$
$$ = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \int_\mathbb{R} dy \ \mathbb{1}^x_{[a,b]} \mathbb{1}^y_{[a,b]} \delta(z-(x-y))=$$
$$ = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \int_\mathbb{R} dy \ \mathbb{1}^x_{[a,b]} \mathbb{1}^y_{[a,b]} \delta(y-(x-z))=$$
$$ = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \ \mathbb{1}^x_{[a,b]} \mathbb{1}^{x-z}_{[a,b]} =$$
The product of the two indicators is equivalent to the conditions:
$$a \leq x \leq b \quad \wedge \quad z+a\leq x \leq z+b.$$
For different values of $z$, these conditions can be reduced to a single inequality, for instance if $0 \leq z \leq b-a$, then the inequality $a+z \leq x $ is "stronger" than $a \leq x$, and the inequality $x \leq b+z$ is "weaker" than $x \leq b$. So for those values of $z$, you have the product of the indicators simplifying to:
$$\mathbb{1}^x_{[a,b]} \mathbb{1}^{x-z}_{[a,b]} = \mathbb{1}^x_{[a+z,b]}, \quad 0 \leq z \leq b-a.$$
Similarly, for another set of values of $z$ (negative $z$, but not too negative), you'll have a different total constraint, and if $z +a \geq b$ or $z+b \leq a$, the two indicators will be totally disjoint and give zero.
To help visualize this, you can draw a picture like this:
$$...---[--*---------]--*---...$$
where the square brackets denote the interval $[a,b]$, and the stars are $a+z$ and $b+z$ respectively.
In your solution you had the wrong numerical factor, $1/4$ instead of a $4$.
Best Answer
We have $$f_Z(z) = \int_{-\infty}^\infty f_X(x)f_Y(z+x) \, dx$$
By doing the substitution of $y=-z+x$ in the second equality below and using the property that $f_x(t)=f_y(t)$ (Note this is different from what you wrote, the argument is the same on both sides of the equality).
\begin{align} f_Z(-z) &= \int_{-\infty}^\infty f_X(x)f_Y(-z+x) \, dx\\ &=\int_{-\infty}^\infty f_X(y+z)f_Y(y) \, dy\\ &=\int_{-\infty}^\infty f_Y(y+z)f_X(y) \, dy\\ &=\int_{-\infty}^\infty f_Y(x+z)f_X(x) \, dx\\ &= f_Z(z) \end{align}