For two independent $U(0,1)$ random variables, the joint density of the maximum and minimum is uniformly distributed on the triangular region with vertices at $(0,0)$,
$(1,0)$ and $(1,1)$. So, yes, given that the maximum has value $k \in (0,1)$,
the minimum is uniformly distributed on $(0,k]$. Similarly, given that the minimum
has value $\ell \in (0,1)$, the maximum is uniformly distributed on $[\ell, 1)$.
Edit added information as requested by OP
Let $W = \max\{X,Y\}$ and $Z = \min\{X,Y\}$. Then, for $a \geq b$,
$$\begin{align}
F_{W,Z}(a,b) &= P\{W \leq a, Z \leq b\}\\
&= P\left(\{W \leq a\}\cap \{Z \leq b\}\right)\\
&= P\left(\{X \leq a, Y \leq b\} \cup \{X \leq b, Y \leq a\}\right)\\
&= P\{X \leq a, Y \leq b\} + P\{X \leq b, Y \leq a\}
- P\{X \leq b, Y \leq b\} )\\
&= F_{X,Y}(a,b) + F_{X,Y}(b,a) - F_{X,Y}(b,b).
\end{align}$$
If one of the steps puzzles you, think about
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$). On the
other hand, if $a < b$, then
$$F_{W,Z}(a,b) = P\{X \leq a, Y \leq a\} = F_{X,Y}(a,a).$$
For jointly continuous random variables $X$ and $Y$, this
gives (upon taking partial derivatives with respect to $a$ and
$b$ that
$$f_{W,Z}(a,b) = \begin{cases}
f_{X,Y}(a,b) + f_{X,Y}(b,a), &\text{if}~ a \geq b,\\
0, &\text{if}~ a < b.
\end{cases}$$
If you think of the joint density $f_{X,Y}$ as a solid sitting
on the plane, just fold it over the diagonal line $a=b$ to
get the joint density $f_{\max,\min}$ of the maximum and
the minimum.
For your instance, the unit cube becomes a right triangular
prism of height $2$ with vertices at $(0,0)$,
$(1,0)$ and $(1,1)$. Thus, the conditional density of
one of the two random variables given the value of the
other, being just a scaled version of the cross-section
of the joint density, is a
uniform density.