The law of total probability

Step 1 (partition $A$ using the $B_i$): define $A_i = A \cap B_i$ for each $i = 1, \ldots, n$. Since the $B_i$ are pairwise disjoint, so are the $A_i$, and since $B_1 \cup \cdots \cup B_n = \Omega \supseteq A$, every outcome of $A$ lies in exactly one $A_i$, so $A = A_1 \cup A_2 \cup \cdots \cup A_n$ with the union disjoint.

Step 2 (add probabilities of disjoint pieces): because probability is additive over pairwise disjoint events, $P(A) = P(A_1) + P(A_2) + \cdots + P(A_n) = \sum_{i=1}^{n} P(A \cap B_i)$.

Step 3 (rewrite each term with the multiplication rule): for each $i$, the multiplication rule gives $P(A \cap B_i) = P(B_i) \cdot P(A \mid B_i)$, which is well defined since $P(B_i) > 0$.

Step 4 (substitute back): replacing every $P(A \cap B_i)$ in Step 2 by $P(B_i) \cdot P(A \mid B_i)$ gives exactly $P(A) = \sum_{i=1}^{n} P(B_i) \cdot P(A \mid B_i)$, as claimed.