Ok first off make sure you understand that $W=X-Y = (\theta + U[-1/2,1/2]) - (\theta + U[-1/2,1/2]) = U[-1/2,1/2]-U[-1/2,1/2]$

Further these are independent so the joint distribution of $X,Y$ is a 2D Uniform variate on $[-1/2,1/2]\times [-1/2,1/2]$

We want $P[W < w] = P[X-Y < w]$.

There are two cases, $w< 0,~0\leq w$

I'll show 3 pictures. The first picture shows when $X-Y < w$ and $w<0$

The beige area is the region that satisfies $P[W < -0.75]$

It should be clear that $x$ ranges from $-1/2$ to some value and that $y$ ranges from some value to $1/2$ The line of the boundary is $Y=X - w$

The $x$ limit occurs when $X-w$ equals the max of $Y$ which is $1/2$, so $X-w = 1/2,~X = 1/2+w$

So the integration limits on $x$ are $-1/2 \to w + 1/2$

Similarly we find the lower limit for $Y$ by noting $X - Y < w \Rightarrow Y > X - w$ and thus for a given $x$, $y > x-w$

Thus the limits on $y$ are $x-w \to 1/2$

The second picture is where $X=Y$

It should be pretty clear $P[W \leq 0] = 1/2$

Now look at $w>0$

Here it's easier to look at $1-P[W > w]$

This corresponds to the purple area.

It should be clear that $x$ goes from some value based on $w$ to $1/2$

And likewise $y$ goes from $-1/2$ to some value based on $w$

$X-Y > w \Rightarrow X > w+Y \geq w -1/2$

and thus the limits on $x$ are $w - 1/2 \to 1/2$

$X-Y > w \Rightarrow Y < X-w \leq 1/2 - w$

and thus the limits on $y$ are $-1/2 \to 1/2 - w$

The coefficient of $1/2$ in front of the last integral is wrong. It should just be 1.

Hopefully you can understand all this. This is a pretty standard problem.