Suppose player $i$ plays the mixed strategy $\mathbb{P}_i(B)= p_i$, and assume for now that the support of $\mathbb{P}_i$ is $\{B,F\}$ (i.e. player 1 plays a fully mixed strategy). For both $B$ and $F$ to be in 1's support, he must obtain the same expected payoff from either strategy (otherwise, he would put all the weight on the strategy with the higher payoff!).
Now, the expected utility of player 1 from playing $B$ is:
$\mathbb{E}[u_1(B,.)] = p_2u_1(B,B)+(1-p_2)u_1(B,F) = 2p_2$
Similarly, the expected utility of player 1 from player in $F$ is:
$\mathbb{E}[u_1(F,.)] = p_2u_1(F,B)+(1-p_2)u_1(F,F) = 1-p_2$
The important point to note here is that the expectation is over the actions of player 2 - since player 1 knows the distribution used by player 2 (in equilibrium), but not the realised action.
You can write the best response function of player 1 as follows:
$BR_1(p_2) = \begin{cases}
0 & \text{ if } 2p_2 < 1-p_2 \\
(0,1) & \text{ if } 2p_2 = 1-p_2\\
1 & \text{ if } 2p_2 > 1-p_2 \\
\end{cases}$
Since we assumed that 1 uses a fully mixed strategy, the $BR_1$ function dicates that this can happen only when $2p^*_2 = 1-p^*_2 \implies p^*_2 = \frac{1}{3}$.
In other words, $p^*_2$ is the unique probability that is consistent with player 1 mixing over both her strategies. Whether this forms an equilibrium is still not clear - for that you need to calculate $BR_2$ (using the same steps) and see if $p^*_1\in(0,1)$. In that case, both players are best responding to each other - and hence playing a Nash Equilibrium.
P.S - for instance, if you found out that $p^*_1 = 1$ (i.e. 1 would like to play pure strategy $B$), then our starting assumption is wrong! So we need to redo the calculation for $p^*_2$.
