Suppose that $X$ and $Y$ are independent random variables. Then, the
the conditional pdf of $Y$ given the value of $X$ is the same as the unconditional pdf of $Y$:
$$f_{Y\mid X}(y\mid X=x) = \frac{f_{X,Y}(x,y)}{f_X(x)} =
\frac{f_X(x)f_Y(y)}{f_X(x)} = f_Y(y).$$
Now consider the computation of $P\{X+Y\leq 2A\}$ via the law of total
probability expression
\begin{align}
P\{X+Y \leq 2A\} &= \int_{-\infty}^\infty P\{X+Y \leq 2A \mid X=x\}
\cdot f_X(x)\, \mathrm dx\\
&= \int_{-\infty}^\infty P\{Y \leq 2A-x \mid X=x\}
\cdot f_X(x)\, \mathrm dx\\
&= \int_{-\infty}^\infty \left[\int_{-\infty}^{2A-x}
f_{Y \mid X=x}(y \mid X = x)\,\mathrm dy\right]
\cdot f_X(x)\, \mathrm dx\\
&= \int_{-\infty}^\infty \left[\int_{-\infty}^{2A-x}
f_{Y}(y)\, \mathrm dy\right]
\cdot f_X(x)\, \mathrm dx
\end{align}
For the special case when $X$ and $Y$ are also identically distributed and
so $f_X(\cdot)=f_Y(\cdot) = f(\cdot)$, and also that $f(\cdot)$ has support $[0,b]$, the above simplifies to
$$P\{X+Y \leq 2A\} = \int_{0}^b \left[\int_{0}^{2A-x}f(y)\, \mathrm dy\right]
\cdot f(x)\, \mathrm dx. $$
There is, to the best of my knowledge, no simple expression relating the value of the above integral to $F(A)$. In particular, the claim in another answer that the value of the integral is $F(b)*F(2A-x)$ is incorrect. Note that $F(b) = 1$ and since the support is $[0,b]$ and
what remains is a function of $x$ but $x$ has been "integrated out" in the double integral.
