= Solution
Integrating the <joint probability density> across $u<v<2u$ gives the <marginal density>
$$
f_U(u)=\int_u^{2u}e^{-u}\,dv=u e^{-u},\qquad u>0.
$$
For every $u>0$, the <conditional probability density> is consequently
$$
f_{V\mid U}(v\mid u)=\frac{f_{U,V}(u,v)}{f_U(u)}
=\frac1u\mathbf1_{\{u<v<2u\}}.
$$
Hence
$$
\boxed{V\mid U=u\sim\operatorname{Unif}(u,2u).}
$$
This also follows from <two independent exponential variables conditioned on their sum>: given $X+Y=u$, the variable $Y$ is uniform on $(0,u)$, and $V=u+Y$. Conditioning on $U=u$ here is defined through the <conditional probability density>, rather than division by $\mathbb P(U=u)$, which is zero. The endpoint $u=0$ lies outside the positive marginal support and needs no specified conditional law.
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