Let $F$ denote the cdf of the uniform distribution over $[0,1]$, the common cdf of $X_1$ and $X_2$. Now, because of the independence of $X_1$ and $X_2$, we have
$$F_{X_1+2X_2}(y)=P(X_1+2X_2<y)=\int_0^1P(x+2X_2<y)\ dx=$$
$$=\int_0^1P\left(Y_2<\frac{y-x}2\right)\ dx=\int_0^1F\left(\frac{y-x}2\right)\ dx.$$
Introducing the new variable $u=\frac{y-x}2$ we get $dx=-2du$, $\frac y2$ for the lower limit, and $\frac{y-1}{2}$ for the upper limit of integration. That is,
$$F_{X_1+2X_2}(y)=2\int^{\frac y2}_{\frac {y-1}{2}}F(u) \ du=\begin{cases}
0&\text{ if }&y<0\\
2\int_0^{\frac y2}u\ du&\text{ if }&0\leq y<1\\
2\int_{\frac {y-1}{2}}^{\frac y2} u\ du&\text{ if }&1\le y\leq2\\
2\int_{\frac {y-1}{2}}^{1} u\ du+2\int_1^\frac y2\ du&\text{ if }&2\le y\leq3\\
1&\text{ if }&y>3.
\end{cases}.$$
Finally
$$F_{X_1+2X_2}(y)=\begin{cases}
0&\text{ if }&y<0\\
\frac14 y^2&\text{ if }&0\leq y<1\\
\frac12y-\frac14&\text{ if }&1\le y\leq2\\
-\frac14y^2+\frac32y-\frac54&\text{ if }&2<y\leq3\\
1&\text{ if }&y>3
\end{cases}.$$
The density is:
$$f_{X_1+2X_2}(y)=\begin{cases}
0&\text{ if }&y<0\\
\frac12y&\text{ if }&0\leq y<1\\
\frac12&\text{ if }&1\le y\leq2\\
-\frac12y+\frac32&\text{ if }&2<y\leq3\\
0&\text{ if }&y>3
\end{cases}.$$
EDIT
Some may find the solution above to be overcomplicated. So let's see what the simple text book solution would be.
The pdf $X_1$, say, $f_{X_1}$ is $1$ over the interval $[0,1]$ and $0$ elsewhere. The pdf belonging to $2X_2$, say $f_{X_2}$ is $\frac12$ over $[0,2]$ and $0$ without it.
So, we simply have to compute the following convolution:
$$f_{X_1+2X_2}(y)=\int_{-\infty}^{\infty}f_{X_1}(y-x)f_{X_2}(x)\ dx.\tag1$$
So far straightforward, indeed. However, when calculating the latter integral, all the boring technical details will come up.
I chose the first solution because in my opinion the boring technicalities - I would dare to say: the same technicalities - emerge closer to and are better explained by the spirit of the original question.
EDIT 2
This is to override my EDIT above.
Let's see how intuitive it can be to compute such a convolution. Consider the following figures
Obviously, if $y<0$ or $y>3$ then the result of the integral of the product is $0$. (see figures (a) and (e))
If $0\leq y\leq 1$ then the integral of the product is increasing as $\frac12y$. (See figure (b).)
If $1<y\leq 2$ then the integral equals $\frac12$. (See figure (c).)
If, however $2<y\leq 3$ the the value of the integral decreases with $y$ from $\frac12$ to $0$. (See figure (d). So for $2<y\leq 3$ the integral equals $-\frac12y+ \frac32.$
Now, is this really simpler? Do we not see the danger in pretending that computing a convolution is always this simple?
Best Answer
I, too, feel it is easier to write the PDF or the CDF by inspection from the graph, but if you wish to stick with convolution for the CDF, your problem is that you are missing a term that represents the cases where $0 \leq X_1 \leq a-2$, and then $X_2$ can be anything (that is, it is certain that the sum is less than $a$, no matter the value of $X_2$). That is, the proper expression is
$$ F_Z(a) = \int_{x_1=a-2}^1 F_{X_2}(a-x_1) \, dx_1 \color{red}{ + \int_{x_1=0}^{a-2} \, dx_1} \qquad 2 \leq a \leq 3 $$
I think if you carry out that integration, you will get the proper answer of
$$ F_Z(a) = -\frac{a^2}{4} + \frac{3a}{2} - \frac{5}{4} \qquad 2 \leq a \leq 3 $$