Determination of (R) and (S) configurations from a Haworth projection
In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.



However, I notice that the question uses (R) and (S) a lot and attempts to argue a lot with it, so a note must be dropped on those stereodescriptors, too. It is important to realise that the (R) and (S) stereodescriptors can change even if the conformation of the carbon in question did not invert or otherwise change conformation. This is due to the way priorities are determined in the CIP system. Let me exemplify this.

First, remember that &#x3b1;-D-glucopyranose, &#x3b2;-D-glucopyranose and the aldehyde form of D-glucose are three different compounds with different physical properties. The &#x3b1; and &#x3b2; forms are diastereomers, more specifically epimers, more specifically anomers of each other and e.g. display different optical rotation. The chain-form of glucose is different altogether. I will construct all further discussion of this answer in a way that it does not matter whether we are dealing with &#x3b1; or &#x3b2;-D-glucopyranose, so you may always assume a mixture.

Now let’s consider the absolute configuration of carbon 3 in both the chain and pyranose forms.


Pyranose


Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbon atoms of the (b) and (c) priorities; initially undistinguishable. Thus, we must go down one level.
Both carbons are attached to the atoms (in order of priority) $\ce{O, C, H}$. Again, this is a tie; we must go down one level.
Both oxygens are attached to a hydrogen; this is identical and the end of the sub-chain. We must go back to the next, lower priority of the former atom.
The two carbon atoms differ: one is the acetal group which expands to $\ce{O, O, H}$. The other is connected to $\ce{O, C, H}$ (carbon 5). Thus, we have resolved the issue.
Going towards lower carbon numbers is the higher priority.


Therefore, carbon $\ce{C{3}}$ has an (S) configuration.
Chain form


Again, oxygen obviously has highest priority.
Equally again, hydrogen obviously has lowest priority.
The two carbons are initially equal.
Both are again attached to $\ce{O(H), C, H}$.
The two subsequent carbons differ again. This time, one is the aldehyde group which expands to $\ce{O, (O), H}$, the other is $\ce{C{5}}$ which expands to $\ce{O, C, H}$. Since a ghost oxygen atom has a higher priority, going up the Fisher projection has the higher priority. 


We arrive at the same (S) conclusion as before.


Now, we will construct the same discussion for carbon $\ce{C{4}}$.


Pyranose


Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbons, initially indistinguishable again. We go down a level.
Both carbons ($\ce{C{3}}$ and $\ce{C{5}}$) expand to $\ce{O, C, H}$. We must go down a level.
One of the two oxygens is connected to a carbon atom. Thus, $\ce{O(C), C, H}$ wins over $\ce{O(H), C, H}$.
Going towards $\ce{C{5}}$ has a higher priority.


Thus, carbon $\ce{C{4}}$ is (S) configured.
Chain form


Bla, oxygen.
Bla, hydrogen.
Blabla, two carbons. Again, we descend a level to $\ce{C{3}}$ and $\ce{C{5}}$.
Both carbons expand to $\ce{O, C, H}$. We descend a level.
Both oxygens are attached to a hydrogen and nothing else. They are indistinguishable.
The two carbons we are now examining are $\ce{C{2}}$ and $\ce{C{6}}$. The former expands to $\ce{O, C, H}$, the latter to $\ce{O, H, H}$. Therefore, Going towards $\ce{C{3}}$ has a higher priority.


Thus, carbon $\ce{C{4}}$ is (R) configured.


Note that while the configuration of $\ce{C{3}}$ did not change, $\ce{C{4}}$’s absolute configuration changed from (R) to (S) upon formation of the pyranose ring. Thus, you cannot rely on absolute configurations to verify a correct Fischer to Haworth transformation for carbohydrates!