Let $Z = 2A + B$ where $A$ and $B$ are uniformly distributed between $[0,1]$. I want to find $f_{A|Z=z}(a|z)$ using cdf. Then:
\begin{equation} F_{A|Z=z}(a|z) = P(A < a) = P(\frac{z-B}{2} < a) = P(z-2a < B) \end{equation}
\begin{equation} F_{A|Z=z}(a|z) = 1-F_B(z-2a) \end{equation}
\begin{equation} f_{A|Z=z}(a|z) = 2f_B(z-2a) \end{equation}
How can I proceed from here?
$$P(A \leq a | z) = \frac{P(A \leq a, Z \in [z-dz,z])}{P(Z \in [z-dz,z])} = \frac{P(A \leq a, 2A+B \in [z-dz,z])}{P(Z \in [z-dz,z])} $$ $$ = \int_{0}^a \frac{P(A \in [s-ds,s], B \in [z-2A-dz,z-2A])}{P(Z \in [z-dz,z])}= \int_{0}^a \frac{P(A \in [s-ds,s], B \in [z-2s+2ds-dz,z-2s])}{P(Z \in [z-dz,z])}$$ Applying Mean Value theorem and assuming continuity of $f_A(x)$ in the interval $[0,a]$, continuity of $f_B(x)$ at in the interval $[z-2a,z]$ and continuity of $f_Z(x)$ at $x = z$: $$= \int_{0}^a \frac{f_A(s) f_B(z-2s) \ ds }{f_Z(z)} $$ Hence : $$f_{A|Z}(a|z)= \frac{f_A(a) f_B(z-2a)}{f_Z(z)} $$.
This does not quite apply to your problem as uniform distribution has two discontinuities in its density function. But where ever the continuity conditions i mentioned holds, the above is the density function. I am hoping that this will help you in general.