I will denote indicator function of event $A$ as $\mathbf 1_{A}(w)$ and denote density of random variable $\phi$ as $\rho_{\phi}(x)$.
Suppose that $\xi$ and $\eta$ are indepent random variables such that $\rho_\xi(x) = \frac{1}{2}e^{-|x|}$, $\rho_\eta(y) = \mathbf 1_{1 \leq y \leq 2}(y)$, so $\xi$ is Laplace random variable and $\eta$ is uniform on the interval $[1,2]$.
I want to find $\rho_{\xi-2\eta}(v)$.
At first I denoted $-2\eta$ as $\zeta$ and found (using CDF of $\eta$) that $\rho_\zeta(z) = \frac{1}{2}\mathbf 1_{-4 < z \leq -2}(z)$.
So now my task is to find $\rho_{\xi + \zeta}(v)$. I was solving it using the formula of convolution of the sum:
$\rho_{\xi+\zeta}(v) = \int_{-\infty}^{+\infty}\rho_{\xi}(u)\rho_{\zeta}(v-u)du = \\ = \int_{-\infty}^{+\infty}\frac{1}{2}e^{-|u|}\frac{1}{2}\mathbf 1_{-4\leq v-u \leq -2}du \stackrel{(1)}{\ne} \frac{1}{2}\int_{max\{0, v+2\}}^{4+v}e^{-u}\mathbf 1_{u \geq 0}du$.
I thought that eqality $(1)$ is true because of evenness of the integrand but I was wrong.
So I broke the integral $\int_{-\infty}^{+\infty}\frac{1}{2}e^{-|u|}\frac{1}{2}\mathbf 1_{v+2\leq u \leq v+4}du$ into two parts: when $u \geq 0$ and when $u < 0$.
As $u \geq 0$, $-4 \leq v < -2$ or $v \geq -2$.
So, when $-4 \leq v < -2$ we have $\frac{1}{4}\int_{0}^{+\infty}e^{-u}\mathbf 1_{v+2\leq u \leq v+4}du = \frac{1}{4}\int_{0}^{v+4}e^{-u}du = \frac{1}{4}(1-e^{-v-4})$
When $ v \geq -2$ we have $\frac{1}{4}\int_{0}^{+\infty}e^{-u}\mathbf 1_{v+2\leq u \leq v+4}du = \frac{1}{4}\int_{v+2}^{v+4}e^{-u}du = \frac{1}{4}(e^{-v-2}-e^{-v-4})$.
As $u < 0$, the $v:$ $-4 \leq v < -2$ or $v < -4 $.
So, when $-4 \leq v < -2$ we have $\frac{1}{4}\int_{-\infty}^{0}e^{u}\mathbf 1_{v+2\leq u \leq v+4}du = \frac{1}{4}\int_{v+2}^{0}e^{u}du = \frac{1}{4}(1-e^{v+2})$
When $ v < -4$ we have $\frac{1}{4}\int_{-\infty}^{0}e^{u}\mathbf 1_{v+2\leq u \leq v+4}du = \frac{1}{4}\int_{v+2}^{v+4}e^{u}du = \frac{1}{4}(e^{v+4}-e^{v+2})$.
When, in the end I got:
$\fbox{$\rho_{\xi+\zeta}(v) = \frac{1}{4}\mathbf 1_{v < -4}(v)(e^{4+v} - e^{2+v}) +\frac{1}{4}\mathbf 1_{-4 \leq v < -2}(v)(2- e^{-v-4} - e^{2+v}) + \frac{1}{4}\mathbf 1_{v \geq -2}(v)(e^{-v-2} - e^{-v-4})$}$
Now are my solution and answer correct?
To simplify let us compute the density $f(x)$ of $X=\xi+Y$ where $Y$ is uniform on $(-1,1)$, namely $f(x)=\frac{1}{4}\int_{-1}^1e^{-|x-y|}dy.$ Because of the symmetry, only the cases $x>1$ and $0<x<1$ are considered: $$f(x)=\frac{1}{2}e^{-x}\sinh 1\ , \ x>1.$$ For $0<x<1$ we have $f=f_1+f_2$ with $$f_1(x)=\frac{1}{4}\int_{-1}^xe^{-x+y}dy=\frac{1}{4}(1-e^{-x-1}),\ f_2(x)=\frac{1}{4}\int_{x}^1e^{x-y}dy=\frac{1}{4}(1-e^{x-1})$$ leading to $$f(x)=\frac{1}{2}-\frac{e^{-1}}{2}\cosh x,\ 0<x<1.$$ The density of $X$ is $f(|x|)$ and the density that you look for is the density of $X-3=V$ which is $f(|v+3|),$ giving a result different of yours.