I am reading the book Stochastic Processes by J. L. Doob and trying to understand the argument that, for any (real-valued) Gaussian process $X = (X_t)_{t\ge0}$, the Markov property is characterized by the condition $$ \mathbb E[X_t \mid X_u : u \le s] = \mathbb E[X_t \mid X_s], \quad \text{a.s., for all } t \ge s. $$ instead of $$ \mathbb P[X_t \in A\mid X_u : u \le s] = \mathbb P[X_t \in A\mid X_s], \quad \text{a.s., for all } t \ge s, A \in \mathcal B(\mathbb R). $$ To this end, we want to show that "the conditional distribution of a Gaussian random process is completely determined by its conditional expectation". To demonstrate this, on page 90, Doob notes that the random variable $X_{t_n} - \mathbb E[X_{t_n} \mid X_{t_1}, \dots, X_{t_{n-1}}]$ is Gaussian with mean zero and independent of $X_{t_1}, \dots, X_{t_{n-1}}$ for any increasing sequence of times. From this, he concludes that "the conditional distribution of $X_{t_n}$ given $X_{t_1}, \dots, X_{t_{n-1}}$ is that of $X_{t_n} - \mathbb E[X_{t_n} \mid X_{t_1}, \dots, X_{t_{n-1}}]$".
I don't quite see what is happening here. What would be the mathematical formulation of the latter claim in quotes? How does this follow from the preceding observations? Together, how would this show that the Markov property is determined by the conditional expectations?
I understand that, for instance, we can derive $$\mathbb P[X_{t_n} - \mathbb E[X_{t_n} \mid X_{t_1}, \dots, X_{t_{n-1}}] \in A\mid X_{t_1}, \dots, X_{t_{n-1}}] = \mathbb P[X_{t_n} - \mathbb E[X_{t_n} \mid X_{t_1}, \dots, X_{t_{n-1}}] \in A]$$ from this independence, but we would have to say something about $$ \mathbb P[X_{t_n} \in A\mid X_{t_1}, \dots, X_{t_{n-1}}]. $$
