We know that if $A$, $B$ and $C$ are mutually disjoint events i.e. if $A \cap B = B \cap C = C \cap A = \emptyset$, then $$\mathbb{P}(A \cup B \cup C) = P(A) + P(B) + P(C)$$
Now consider two events $E_1$ and $E_2$ that are not mutually disjoint events i.e. if $E_1 \cap E_2 \neq \emptyset$ and we want to evaluate $P(E_1 \cup E_2)$.
From the figure, it is apparent that the intersection $E_1 \cap E_2$ is counted twice: once as part of $E_1$ and once as part of $E_2$. Hence, it needs to be subtracted off. For a more rigorous derivation, see below the figure.

The idea is to split $E_1 \cup E_2$ into three disjoint sets as follows. $$E_1 \cup E_2 = (E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1)$$
Note that $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ and $(E_2 \backslash E_1)$ are mutually disjoint sets. Hence, we have that $$P(E_1 \cup E_2) = P((E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1))\\ = P(E_1 \backslash E_2) + P(E_1 \cap E_2) + P(E_2 \backslash E_1)$$ Call the above equation $\star$.
Now note that $$(E_1 \backslash E_2) \cup (E_1 \cap E_2) = E_1$$ Since $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ are mutually disjoint sets, we have that $$P(E_1 \backslash E_2) + P(E_1 \cap E_2) = P(E_1)$$ Hence, $$P(E_1 \backslash E_2) = P(E_1) - P(E_1 \cap E_2)$$ Similarly, since $$(E_1 \cap E_2) \cup (E_2 \backslash E_1) = E_2$$ are mutually disjoint sets, we have that $$ P(E_1 \cap E_2) + P(E_2 \backslash E_1) = P(E_2)$$ Hence, $$ P(E_2 \backslash E_1) = P(E_2) - P(E_1 \cap E_2)$$ Now plug in for $P(E_1 \backslash E_2)$ and $P(E_2 \backslash E_1)$ in $\star$, to get what you want.