Nested implications vs. two implications with conjunction Assuming I want to translate these statements into propositional logic:

If it rains, the street gets wet.
If the street is wet, it's slippery.

A straightforward translation is:
$$    (R \Rightarrow W) ∧ (W \Rightarrow S)$$
Where

*

*$R$: It rains

*$W$: The street is wet

*$S$: The street is slippery

But what about this translation:
$$(R \Rightarrow W) \Rightarrow S$$
I can see that the truth tables of $(R ⇒ W) ⇒ S$ and $(R ⇒ W) ∧ (W ⇒ S)$ are different but I'm having trouble finding an intuitive explanation of why one of them is a valid translation and the other one is incorrect.
How do those two formulas differ semantically?
 A: Just because when it rains the street gets slippery does not mean that the street is slippery now (maybe it hasn't rained yet). 
What I think you find intuitive is
$$(R \land (R \Rightarrow W) \land (W \Rightarrow S)) \Rightarrow S,$$
i.e., that if it rains and rain implies wetness implies slipperiness, then it's also slippery. Note that you have to assume that it rains in order for this to be true.
Note that $R \Rightarrow W$ stands for (whenever it rains it's wet), but this statement alone does not imply the statement $S$ (it's slippery). It only implies $S$ when combined with $R$. So $(R \Rightarrow W) \Rightarrow S$ is not true (because sometime's its not slippery, even though when it rains then it's wet).
A: 
The first one is :
$$(R\to W)∧(W\to S)$$
$1.$If it rains, the street gets wet.
$2.$If the street is wet, it's slippery.

The second one is:
$$(R\to W)\to S\equiv (\neg R\to S)\land (W\to S)$$
$1.$If it's not rains, it's slippery.
$2.$If the street is wet, it's slippery.
We also have the following by taking contrapositive of $\neg R\to S$ $$(R\to W)\to S\equiv (\neg S\to R)\land (W\to S)$$
$1.$If it's not slippery, it rains.
$2.$If the street is wet, it's slippery.
The difference should be clear now
A: In the intended interpretation ( assuming the street we are talking about is a street with open sky) the formula : 
(R--> W) & (W-->S) 
is true. 
In the actual world ( with its actual physical properties) the fact that it rains " implies" ( though not in the logical sense) that the street is wet and the fact that this street is wet also " implies" that it is slippery. 
Indeed, there is no possible ( physical) situation in which (1) it rains and the street is not wet and no possible (physical) situation in which (2) the street is wet and is not slippery. 
On the other hand, still in the intended interpretation, the formula : 
( R --> W)  ---> S 
can be true or false. 
Indeed, the true fact that " if it rains the street is wet " does not mean that the street is now slippery. 
The antecedent of this conditional  formula is true ( in the actual physical world). So, in order the formula to be true, S would have to be actually  true. ( See the truth table of the " if ... then" operator). 
But nothing guarantees that "S" is true : in the actual world, S can be true or false. 
So the whole conditional is not true in all possible ( physical) cases. 

Other example: 
If Bob is a pediatrician, then Bob is a doctor & if Bob is a Doctor then Bob is educated. 
This formula, though not formally valid, is logically true. It is logically impossible (1) that Bob is a pediatrician without being a Doctor and (2) that Bob is Doctor without being educated. 
This is different from saying that : 
if  ( Bob is a pediatrician implies Bob is a doctor) then Bob is educated, 
or , in symbols : (P --> D) --> E. 
The logical and general fact Bob ( or any other person) could not be a pediatrician without being a doctor does not imply at all that this particular guy, Bob, is actually a pediatrician, a doctor, or an educated man. 
A: Let's coinsider the case where all three statements involved are actually the same statement. For example, suppose we have:
$P$: God Exists
OK, then $(P \to P) \land (P \to P)$ is of course the conjunction of two tautologies, and hence is a tautology itself: yes, of course it is true that if God exists, then God exists, (and saying it twice does not take anything away from that!)
However, now look at $(P \to P) \to P$
This would translate back as: "If it is true that God exists of God exists, then God exists". But of course course God exists if God exists, and so the antecedent of the outside conditional is true, meaning that this statement is equivalent to $P$, and thus ends up saying "God exists". Well, that may be true, but is clearly not a tautology!
