THE PROBLEM
HEURISTIC SOLUTION
Since $X \sim$ Unif$(-a,a)$, the restriction that $X+Y=u_0$ automatically imposes the following restriction on $Y$ :
$$Y \in (u_0-a,u_0+a) \tag{1}$$
Hence the required probability, which is scaled with respect to the restriction $(1)$, is
$$\frac{f(y)}{P\left\{Y \in (u_0-a,u_0+a)\right\}} \times I\left\{Y \in (u_0-a,u_0+a)\right\}$$
which reduces to the desired quantity. The problem is to write this in a rigorous manner.
ATTEMPTED EFFORT FOR A RIGOROUS SOLUTION
$$F_{Y|X+Y}(y|u_0):= \lim_{\epsilon \downarrow 0} P\left\{Y \leq y | X+Y \in (u_0-\epsilon, u_0+\epsilon)\right\}$$
The quantity in the right hand side is basically
$$\lim_{\epsilon \downarrow 0} \frac{P\left\{Y \in (-\infty,y] \cap X+Y \in (u_0-\epsilon, u_0+\epsilon)\right\}}{P\left\{X+Y \in (u_0-\epsilon, u_0+\epsilon)\right\}} \tag{2}$$
This quantity should reduce to : $$\frac{F(y)}{F(u_0+a)-F(u_0-a)} \cdot I\left\{u_0-a<y<u_0+a\right\} \tag{3}$$
Any help regarding showing that the expression in $(2)$ is equal to the expression in $(3)$ would be much appreciated. Also, if there is any short-cut method that achieves the main goal without using the rigorous definition of conditional probability (where the conditioning variable is continuous), I'd be grateful if someone is kind enough to share that.