The reflection principle says that for the Markov process $(B_t,\mathcal{F}_t,P_x)$ associated with Brownian Motion it is satisfied that
$P_0(\text{max}_{s\leq t} B_s \geq a) =2P_0(B_t\geq a)$
What is the interpretation behind it? Why is it called "Reflection Principle"? Is there something that gets reflected?
For $a>0$ denote by
$$\tau_a :=\tau_a^B:= \inf\{t>0; B_t = a\}$$
the first time the Brownian motion hits the line $y=a$.The reflection principle, which you stated in your question, is then equivalent to
$$\mathbb{P}(\tau_a \leq t) = 2 \mathbb{P}(B_t \geq a).$$
The key to prove this equation is the following identity:
$$\mathbb{P}(\tau_a \leq t, B_t < a) = \mathbb{P}(\tau_a \leq t, B_t>a). \tag{1}$$
Indeed, once we have this identity we get
\begin{align*} \mathbb{P}(\tau_a \leq t) &= \mathbb{P}(\tau_a \leq t, B_t < a) + \mathbb{P}(\tau_a \leq t, B_t > a) \\ &= \mathbb{P}(\tau_a \leq t, B_t \geq a) + \mathbb{P}(\tau_a \leq t, B_t\geq a) = 2\mathbb{P}(B_t \geq a). \end{align*}
Equation $(1)$ is the point where the reflection comes into play. Define a new process
$$W_t :=\begin{cases} B_t, & t \leq \tau_a, \\ 2 B_{\tau_a}-B_t, & t>\tau_a \end{cases}$$
Up to time $\tau_a$ the process is just the original Brownian motion and for $t>\tau_a$ we reflect the Brownian motion at the horizontal line $y=a$.
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It turns out that $(W_t)_{t \geq 0}$ is again a Brownian motion. In particular,
$$\mathbb{P}(\tau_a^W \leq t, W_t < a) = \mathbb{P}(\tau_a^B \leq t, B_t < a). \tag{2}$$
By the definition of $(W_t)_{t \geq 0}$ as a reflection of $(B_t)_{t \geq 0}$, we have $\tau_a^W = \tau_a^B$ and $$\{\tau_a^W \leq t,W_t < a\} = \{\tau_a^B \leq t, B_t>a\}.$$ Hence, $(2)$ is equivalent to $$\mathbb{P}(\tau_a^B \leq t, B_t>a) = \mathbb{P}(\tau_a^B \leq t, B_t<a)$$ which is, in turn, equivalent to $(1)$.
In this sense, the reflection principle is a direct consequence of the fact that the reflected process $(W_t)_{t \geq 0}$ is a Brownian motion - justifying the name of the result.