The probability of the union of two events Does the union of two events exclude that those events may occur together?  In other words, is it the same as an exclusive or?
$p(E_1 \cup E_2) = p(E_1) + p(E_2) - p(E_1 \cap E_2)$
The reason I ask is because the above equation explicitly subtracts the probability of the events occurring together.  
 A: 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.
A: The formula you give is the correct way to determine $\Pr(E_1 \cup E_2)$ (often referred to as the inclusion-exclusion principle). The union of the two events, however, does include outcomes occurring in both events. Formally,
$$
E_1 \cup E_2 = \{\omega \mid \omega \in E_1 \text{ (inclusive) or } \omega \in E_2 \}.
$$
The reason we subtract $\Pr(E_1 \cap E_2)$ in the formula you give is because outcomes occurring in the intersection would otherwise be counted twice. To see this, it is easier to just think of sets. Is it true that
$$
|A \cup B| = |A| + |B|?
$$
No. For a simple counterexample to this claim, let $A = B = \{1\}$. We know $A \cup B = \{1\}$, so $|A \cup B| = 1$, while $|A| + |B| = 2$. The reason is because the lefthand side counts the element $1$ a single time, while it is counted twice on the righthand side (once because it belongs to $A$ and once because it belongs to $B$). An analogous argument applies to probabilities.
