Your question can be reframed as: why the statement (a) $X_t \mid X_{t - 1} \sim N(\sqrt{1 - \beta_t}X_{t - 1}, \beta_t I)$ is equivalent to the statement (b) $X_t$ has the representation $X_t = \sqrt{1 - \beta_t}X_{t - 1} + \sqrt{\beta_t}Z_t$, where $Z_t \sim N(0, I)$ is independent of $X_{t - 1}$?

While Thomas Lumley has given you a heuristic answer, I agree that you are absolutely entitled to request a more rigorous argument.

The direction $(b) \Rightarrow (a)$ is an immediate application of the theorem below (Lemma 8.7 in Foundations of Modern Probability (3rd edition) by Olav Kallenberg):

For any random elements $\xi \perp \eta$ in $S, T$ and a measurable function $f: S \times T \to U$, where $S, T, U$ are Borel, define $X = f(\xi, \eta)$. Then $\mathcal{L}(X \mid \xi) = \mathcal{L}\{f(x, \eta)\}|_{x = \xi}$ a.s.

Here “$\mathcal{L}(X)$ / $\mathcal{L}(X \mid \xi)$” denotes the law (i.e., distribution) of $X$ / conditional law of $X$ given $\xi$. To apply this result, simply let $\xi = X_{t - 1}, \eta = Z_t$, and $X = X_t = \sqrt{1 - \beta_t}X_{t - 1} + \sqrt{\beta_t}Z_t$. If you are interested in the proof to this lemma, check the reference by yourself.

For the direction $(a) \Rightarrow (b)$, it suffices to prove the conditional distribution of $Z_t := \beta_t^{-1/2}(X_t - \sqrt{1 - \beta_t}X_{t - 1})$ given $X_{t - 1}$ is $N(0, I)$, hence it is independent of $X_{t - 1}$. To this end, we can apply Theorem 8.5(ii) in the same reference (which concerns with the disintegration of the joint distribution of $(X_{t - 1}, X_t)$):

Theorem (conditional distributions, disintegration) Let $\xi, \eta$ be random elements in $S, T$, where $T$ is Borel. Then $\mathcal{L}(\xi, \eta) = \mathcal{L}(\xi) \otimes \mu$ for a probability kernel $\mu: S \to T$, where $\mu$ is unique a.e. $\mathcal{L}(\xi)$ and satisfies

1. $\mathcal{L}(\eta \mid \xi) = \mu(\xi, \cdot)$ a.s.,
2. $E\{f(\xi, \eta) \mid \xi\} = \int \mu(\xi, dt)f(\xi, t)$, a.s., $f \geq 0$.

By this theorem, for a Borel set $H \in \mathscr{R}^n$, we have (where the notation $aH + b$ denotes the set ${ax + b: x \in H}$): $$ \begin{align*} &\ P(Z_t \in H \mid X_{t - 1}) \\ =&\ P\left(X_t \in \sqrt{\beta_t}H + \sqrt{1 - \beta_t}X_{t - 1} \mid X_{t - 1}\right) \\ =&\ \int_{\sqrt{\beta_t}H + \sqrt{1 - \beta_t}X_{t - 1}} (2\pi\beta_t)^{-n/2}\exp\left(-\frac{\beta_t^{-n}}{2}(x - \sqrt{1 - \beta_t}X_{t - 1})^\top(x - \sqrt{1 - \beta_t}X_{t - 1})\right)dx \\ =&\ \int_H (2\pi)^{-n/2}\exp\left(-\frac{1}{2}z^\top z\right)dz. \end{align*} $$

where the last step uses the transformation $z = \beta_t^{-n/2}(x - \sqrt{1 - \beta_t}X_{t - 1})$ and the property of the Lebesgue integration. The last expression shows that the conditional distribution of $Z_t$ given $X_{t - 1}$ is exactly $N(0, I)$. This completes the proof.