Intuition about a logic equivalence It is fairly easy to show that $(P\wedge Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R)$. 
However I am having trouble having an intuition about this equivalence. My intuition is that if both hypotheses $P$  and $Q$ are required to prove $R$, neither one by itself is sufficient to prove $R$. 
Example 1: If I am very talented at basketball ($P$) and I am a man (Q) then I have a chance to play in NBA ($R$). 
Example 2: If $p>2$ and $p$ is prime then $p$ is odd. However neither hypotheses by themselves imply that $p$ is an odd number.   
Certainly my intuition is very wrong, could anyone give me perhaps more insight?   
 A: The confusion here isn't with implication, but with your implicit use of a universal quantifier.
Look at your Example 2, but write out the quantifier (I'll assume that quantifiers are taken to quantify over the natural numbers):
$$(\forall p)\big((p\gt 2 \,\wedge\, p\text{ is prime}) \to p\text{ is odd}\big). \tag{1}$$
This means that every natural number that is both greater than 2 and prime is, in fact, odd.  (This is true, of course.)
By your propositional logic equivalence, it's correct to say that (1) is equivalent to:
$$(\forall p)\big((p\gt 2 \to p\text{ is odd}) \lor (p\text{ is prime }\to \;p\text{ is odd})\big),\tag{2}$$
although (2) is a strange, hard-to-interpret statement that nobody would write in that way.
This is not, however, equivalent to:
$$\big((\forall p)(p\gt 2 \to p\text{ is odd})\big)\lor\big((\forall p)(p\text{ is prime} \to p\text{ is odd})\big)\tag{3}$$
nor is it equivalent to
$$(\forall p)\big((p\gt 2 \,\lor\, p\text{ is prime} )\to p\text{ is odd})\big).\tag{4}$$
(3) means: Either (a) every natural number greater than 2 is odd or (b) every prime number is odd (or both).
(4) means: Every natural number that is either greater than 2 or prime (or both) is odd.
Both (3) and (4) are false — and, more to the point, they clearly aren't equivalent to (1) and to (2), nor does the propositional logic equivalence say that they should be.

So, in propositional calculus it's true that $(P\wedge Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R),$ and it's therefore true in first-order logic that
$$(\forall x)\big((P(x) \wedge Q(x))\to R(x)\big)$$
is equivalent to
$$(\forall x)\big((P(x)\to R(x))\lor(Q(x)\to R(x)) \big).$$
But this isn't equivalent to
$$(\forall x)\big(P(x)\to R(x)\big) \lor (\forall x)\big(Q(x)\to R(x)\big),$$
for example.  In general, $(\forall x)(A(x)\lor B(x))$ isn't equivalent to $(\forall x)A(x) \lor (\forall x)B(x).$
A: Mitchell Spector's answer explained very well what's wrong with your examples. Let me try to explain intuitively why $(P\land Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R).$ (Since this is an intuitive explanation, I feel absolved from any need to be rigorous or correct.)
Let me identify a proposition with its truth value, i.e., $P=1$ if $P$ is true, $P=0$ if $P$ is false. Then $P\to R$ is true just in case $P\le R,$ i.e., "$R$ is at least as true as $P$ is." Since $P\land Q=\min(P,Q),$ the equivalence
$$(P\land Q)\to R\iff(P\to R)\lor(Q\to R)$$
just says that
$$\min(P,Q)\le R\iff P\le R\text{ or }Q\le R$$
which is a simple arithmetical fact, nothing unintuitive about it.
