So EK says EWG tend to deshield hydrogens and the peak tends to be downfield to the left (they refer to this as lower magnetic field strength) while EDG tend to increase shielding and increase the required field strength (moving the peak upstream to higher magnetic field stength).
On the other hand, BR seems to say differently.
Take the case of 3-pentanone. BR says the CH2 group adjacent to the carbonyl group feels the magnetic field of the electrons on a neighboring oxygen and thus requires a stronger external magnetic field so the peak is shifted downfield.
Seems to me like BR says downfield indicates stronger magnetic field. In contrast, I think EK would say this occurs because the EWG (the carbonyl) deshields the alpha hydrogens resulting in the downfield shift.
I'm not sure which explanation to follow.
My main concern is in the case of ester we have O=C-O-CH3 the shift of the hydrogen is shifted downfield around 3-4 ppm, whereas the alpha carbon of a carbonyl O=C-CH3 is 1-2 ppm. I don't understand the logic why.
There isn't a disagreement here; it's a different perspective. The problem is that NMR shift values take multiple factors into consideration (EDG/EWG impact, electronegativity of nearby groups, and anisotropy), so there is not always a clear answer (as you point out with the ester example). If you go by the EDG vs. EWG apporach endorsed by EK, then the ester will not make sense. If you go by the local B field approach used by BR, then benzene will not make sense. At some point you'll need to decide exactly how much depth you want to know this material, and what ambiguities you can accept and live with. Both approaches will work for the MCAT, and that's ultimately what matters.
Both explanations (EK and BR) point to downfield for the alpha protons of a ketone, so that's the end result you should hang onto. The EK explanation is in line and tone with what you'll find in most undergrad textbooks. The BR explanation is a visual explanation simplifed for MCAT style questions. The reality is that the two explanations both only tell part of the story, but they do agree for simple cases. To answer your ester question, I prefer the visulization of the local B field due to the electron rich oxygen of the alkoxy group requiring a stronger external field to overcome its presence.
If you want to get into the nuts and bolts (go beyond the scope of the MCAT), then consider two things. First, the ppm values on the delta scale (commonly used) actually run opposite of applied field strength (which is measured more correctly on the tau scale) and second exactly what shielded and deshielded mean in terms of orientation of the local B fields of functional groups on the molecule and the anisotropy effect.
If you consider a typical system of 14,100 Gauss, then the RF for a unshielded proton is 60 MHz. A common NMR machine maintains the frequency at 60 MHz (or whatever value the given machine uses--they vary from machine to machine) and varies the applied B field. The delta value is the (shift in Hz)/(machine frequency in MHz). The tau value is found as 10 - delta. The tau scale has TMS at 10 ppm and aldehydes around 0-1. The delta scale has TMS at 0 and aldehyde protons 9-10. The tau value correlates to applied B field strength.
A good example to consider is octadecacyclo-1,3,5,7,9,11,13,15,17-nonene (not perfect IUPAC to describe an 18-carbon ring with alternating conjugated pi-bonds). It has twelve outer Hs with delta values of 8.9 Hz and six inner Hs with delta values of -1.4 Hz. All are technically on the same amount of EDGs, but have drastically differing values. This is attributed to anisotropy, which is best understood by visualization. The point is that to fully get NMR, you need to go into depth with each system and consider the impact of each factor. No watered-down method is perfect, but the detailed approach is cumbersome and beyond what you need for the MCAT. So, accept that the EK method and BR method are not identical in their approach, but both are fine for the MCAT. Use the one that sticks best for you.
