How to describe the coupling between $\mu$ and $\nu$? Here is the definition of the coupling:

In measure theory, coupling μ and ν means constructing a measure π on X ×Y such that π admits μ and ν as marginals on X and Y respectively.
Here is the definition of deterministic coupling:

 A: Quick answer that summarizes my above comments:
Any two random variables $X$ and $Y$ on the same probability space can also be viewed as a random vector $(X,Y)$ that itself already defines a coupling between the distribution of $X$ and the distribution of $Y$. If it just so happens that $Y=T(X)$ for some measurable function $T$ then $(X,Y)$ is in fact a deterministic coupling.
In your example $X \sim Uniform([0,1])$ and $Y=T(X)$ for some measurable $T:[0,1]\rightarrow [a,b]$ (given some real numbers $a,b$ with $a<b$), then indeed $(X,Y)$ is already a deterministic coupling between the distribution $\mu$ of $X$ and the distribution $v$ of $Y$.  Specifically
\begin{align}
&\mu(A) = P[X \in A] \quad \forall A \in \mathcal{B}([0,1])\\
&v(B) = P[T(X) \in B] \quad \forall B \in \mathcal{B}([a,b])
\end{align}

On the reverse situation: Given two distributions $\mu$ and $v$, there might be many ways of constructing a random vector $(X,Y)$ such that $X$ has distribution $\mu$ and $Y$ has distribution $v$.  The simplest way is to impose independence between $X$ and $Y$ via the product measure. However, we can then ask if there is a deterministic coupling (so that $X$ and $Y$ are not necessarily independent). The answer is "not necessarily."  For example if $(X,Y)$ is a random vector where $X$ takes values in the set $\{0,1,2\}$ equally likely, and $Y$ takes values in the set $\{0,1\}$ equally likely, it is not possible to write $X=T(Y)$ or $Y=T(X)$ for some function $T$.
A key interesting result on stochastic coupling between random variables is this: Let $X:\Omega_1\rightarrow\mathbb{R}$ and $W:\Omega_2\rightarrow\mathbb{R}$ be random variables on potentially different probability spaces $(\Omega_1, \mathcal{F}_1, P_1)$ and $(\Omega_2, \mathcal{F}_2, P_2)$ that satisfy
$$P_1[X>x]\leq P_2[W>x]\quad \forall x \in \mathbb{R} \quad (*)$$
Then there is a random variable $Y:\Omega_1\rightarrow \mathbb{R}$ (on the same probability space as $X$) such that  $X(\omega)\leq Y(\omega)$ for all $\omega \in \Omega_1$, and $Y$ has the same distribution as $W$ (so $P_1[Y>y]=P_2[W>y]$ for all $y \in \mathbb{R}$). If (*) holds we say that $X$ is stochastically less than or equal to $W$.
