The Borel $\sigma$-algebra of the set of probability measures

One direction is covered by Alexander Shamov's answer or my comment above. Here is the other direction.

Let $\mathcal{F}$ be the smallest $\sigma$-algebra such that, for every Borel $A \subset X$, $\mu \mapsto \mu(A)$ is $\mathcal{F}$-measurable. As mentioned, we have shown that $\mathcal{F} \subset \mathcal{B}(M(X))$.

Note further that for any bounded Borel $f : X \to \mathbb{R}$, the map $\mu \mapsto \int f\,d\mu$ is $\mathcal{F}$-measurable. (This is clear when $f$ is simple. Otherwise, approximate $f$ by a bounded sequence $f_n$ of simple functions and note that by dominated convergence, $\int f\,d\mu = \lim_{n \to \infty} \int f_n\,d\mu$, so that $\mu \mapsto \int f\,d\mu$ is a pointwise limit of $\mathcal{F}$-measurable functions.)

In particular, this is true when $f$ is continuous. So for any continuous $f$ and $a < b \in \mathbb{R}$, we have $$U_{f,a,b} := \left\{\mu : a < \int f\,d\mu < b\right\} \in \mathcal{F}.$$ But the collection of all $U_{f,a,b}$ is a sub-basis for the the weak-* topology. In particular, every weak-* open set in $M(X)$ is a union of finite intersections of such sets. Since $M(X)$ is second countable, we may say every open set is a countable union of finite intersections of $U_{f,a,b}$ sets, and hence every open set is in $\mathcal{F}$. So $\mathcal{B}(M(X)) \subset \mathcal{F}$ and your characterization is proved.

Consider the set $\mathcal A$ of measurable subsets $A \subset X$, such that $\mu \mapsto \mu[A]$ is measurable. Obviously, $\mathcal A$ contains all open sets.

It's also easy to see that $\mathcal A$ contains an algebra of sets - say, sets $A \subset X$, such that their indicator is a pointwise limit of a sequence of continuous functions. Indeed, if $f_n \to \mathsf{1} [A]$ and $0 \le f_n \le 1$ (which can always be assumed, since we can replace $f_n$ by $\max(\min(f_n, 0), 1)$) then the function $\mu \mapsto \mu[A]$ is the pointwise limit of $\mu \mapsto \intop f_n d\mu$.

A similar argument to the above shows that $\mathcal{A}$ is closed under sequential limits of sets. And anything that contains an algebra and is closed under sequential limits contains a $\sigma$-algebra (I would call that a version of the monotone class theorem). Therefore, $\mathcal A \supset \mathcal{B} (X)$.