Why does an increase in alkyl group substitution cause downfield shift H'NMR?
The short answer to this question is that C is more electronegative than H (H 2.2, C 2.5), and therefore one would expect a trend in increasing chemical shift with increasing carbon substitution.

The longer answer....

As you point out, alkyl groups are electron donating, and you are on the right track in surmising that alkyl groups are less electron donating than protons. Remember that all shielding is relative. Looking at shielding effects at the &alpha;-substituent;


a bare proton nucleus (H+) would come at &delta;35, 
H-H comes at &delta;7.4, and 
H-CH3 comes at &delta;0.23. 


A clear upfield shift due to the electron-donating alkyl group. So, we need to look deeper at the effect of &beta;-substituents:


H-CH3 (&delta;0.23), 
H-CH2CH3 (&delta;0.86), 
H-CH(CH3)2 (&delta;1.33), and 
H-C(CH3)3 (&delta;1.56).


This longer answer involves looking at the magnetic anisotropy around the observed nucleus, and questioning some age-old depictions of the C-C bond. Historically, the C-C magnetic anisotropy was depicted as below, with shielding at the side, and deshielding along the ends. Have a look in your NMR textbook, and you'll probably find this. Taken in isolation, this model conforms with the observed trend of increasing shift with increasing substitution with C.



However, this model contradicts somewhat the notion that alkyl groups are electron donating, and indeed what is observed with &alpha;-substitution. And with good reason. Recent studies by Baranac-Stojanovic (DOI: 10.1002/chem.201204267) show that this anistropy model should be reversed, and that there are shielding zones at the end of the C-C bond. Why then do we observe the increase in chemical shift with increased C substitution? Again, it is because the shielding anisotropy of the contributing C-C bond component is considerably less than contributions from the geminal C-H anisotropy contributions.