The question is a bit vague, and the answer below may not be what you want. I assume that by the union of those spaces you mean disjoint union. Here is one quite general answer whn one can equip such an union with a structure of vector bundle. Assume that $\Omega$ is a domain in a fixed Euclidean space $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. Suppose $(\Psi_s)_{s\geq 0}$ is a smooth family of diffeomorphisms $\Psi_s:\bR^n\to\bR^n$ and $\Omega_s=\Psi_s(\Omega)$. For each $s\geq 0$ denote by $J_s: \Omega_s\to (0,\infty)$ the Jacobian of $\Psi_s$, i.e., the absolute value of the determinant of the differential of $\Psi_s$. For any $s$ we have a natural Hilbert space isomorphism,

$$T_s: H_s:=L^2(\Omega_s)\to H:=L^2(\Omega),\;\; L^2(\Omega_s)\ni f \mapsto J_s^{\frac{1}{2}} \Psi_s^* (f)\in L^2(\Omega). $$

Indeed

$$ \int_\Omega \bigl( \; J_s^{\frac{1}{2}} \Psi_s^* (f)\bigr)^2= \int_\Omega J_s \Psi_s(f^2) dV= \int_{\Omega_s} f^2 dV. $$

where $dV$ denote the Euclidean volume element. We now have a bijection $\newcommand{\T}{\mathscr{T}}$ $\newcommand{\H}{\mathscr{H}}$

$$ \T: \H:=\bigsqcup_{s\geq 0} H_s =\bigcup_{s\geq 0} \{s\}\times H_s \to [0,\infty)\times H, $$

$$ \{s\} \times H_s\ni (s, f) \mapsto (s, T_s f)\in [0,\infty)\times H. $$

There exists a unique topology of $\H$ which makes $\T$ a homeomorphism onto the space $[0,\infty)\times H$ equipped with the product topology.

The question is a bit vague, and the answer below may not be what you want. I assume that by the union of those spaces you mean disjoint union. Here is one quite general answer when one can equip such an union with a structure of vector bundle. Assume that $\Omega$ is a domain in a fixed Euclidean space $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. Suppose $(\Psi_s)_{s\geq 0}$ is a smooth family of diffeomorphisms $\Psi_s:\bR^n\to\bR^n$ and $\Omega_s=\Psi_s(\Omega)$. For each $s\geq 0$ denote by $J_s: \Omega_s\to (0,\infty)$ the Jacobian of $\Psi_s$, i.e., the absolute value of the determinant of the differential of $\Psi_s$. For any $s$ we have a natural Hilbert space isomorphism,

$$T_s: H_s:=L^2(\Omega_s)\to H:=L^2(\Omega),\;\; L^2(\Omega_s)\ni f \mapsto J_s^{\frac{1}{2}} \Psi_s^* (f)\in L^2(\Omega). $$

Indeed

$$ \int_\Omega \bigl( \; J_s^{\frac{1}{2}} \Psi_s^* (f)\bigr)^2= \int_\Omega J_s \Psi_s(f^2) dV= \int_{\Omega_s} f^2 dV, $$

where $dV$ denote the Euclidean volume element in $\bR^n$. We now have a bijection $\newcommand{\T}{\mathscr{T}}$ $\newcommand{\H}{\mathscr{H}}$

$$ \T: \H:=\bigsqcup_{s\geq 0} H_s =\bigcup_{s\geq 0} \{s\}\times H_s \to [0,\infty)\times H, $$

$$ \{s\} \times H_s\ni (s, f) \mapsto (s, T_s f)\in [0,\infty)\times H. $$

There exists a unique topology on $\H$ which makes $\T$ a homeomorphism onto the space $[0,\infty)\times H$ equipped with the product topology.