In the above context, given a Brownian motion B started from a point $x$ in a bounded domain $D$, my notes define the solution to the Dirichlet problem with boundary value $f$ on $\partial D$ to be the function
$$ S(x) = E_x(f(B_T))$$
where $T$ is the exit time of $B$ from $D$. $E_x$ is supposed to indicate the expectation w.r.t. the probability measure under which $B$ started from $x$ is a Brownian motion, but how can I make sense of this? The definition of $P_x$ seems so indirect to me that I cannot fathom this probability measure, nor be confident that I understand correctly. Can somebody clarify the situation in the most direct manner possible?
Your question is just about the definition of Brownian Motion as a probability measure.
Let's work in $\mathbb{R}^2$, and fix some $x \in \mathbb{R}^2$ (you can pretend it's the origin). Let $\Gamma_x = \{\gamma : [0,\infty) \to \mathbb{R}^2 \hspace{1mm} | \hspace{1.5mm} \gamma \text{ is continuous and has } \gamma(0)=x\}$. We define a probability measure $\mathbb{P} = \mathbb{P}_x$ on $\Gamma_x$; the sigma-algebra is the Borel one.
Let $E$ be a Borel subset of $\Gamma_x$. We now define $\mathbb{P}(E)$. We will use some external source of randomness (this hopefully is fine with you). For each $n \ge 1$, let $\gamma_n : \mathbb{Z}^{\ge 0} \to \mathbb{R}^2$ be the random function defined by $\gamma_n(0) = x$, and for $t \ge 1$, $\gamma_n(t+1) := \gamma_n(t)+\sigma_t$, where $\sigma_t$ is chosen uniformly at random from $\{(-1,0),(0,1),(0,-1),(0,1)\}$. Extend $\gamma_n$ to $\mathbb{R}^{\ge 0}$ by defining $\gamma_n(t+\delta) := \gamma_n(t)+\delta(\gamma_n(t+1)-\gamma_n(t))$ for $t \in \mathbb{Z}^{\ge 0}$ and $\delta \in [0,1)$. Then each $\gamma_n$ is continuous with $\gamma_n(0) = x$. Therefore, $\gamma := \sum_{n=1}^\infty \frac{\gamma_n}{2^n}$ is a continuous function with $\gamma(0) = x$. We define $\mathbb{P}(E)$ to be the probability that $\gamma \in E$.
The point is that if you choose an element $\gamma \in \Gamma_x$ randomly according to $\mathbb{P}$, what you get corresponds to your intuition of what Brownian Motion starting at $x$ is; namely, a "random walk" starting at $x$.