Why do we want to compute derived functors via forgetful functors though easier? This is a statement made in Weibel, Homological Algebra, Chpt 2, right after Exercise 2.4.2. Assume $F$ is right exact functor and thus it allows left derived functors. 
"Forgetful functors such as $mod_R\to Ab$ are often exact, and it is often easier to compute the derived functors of $UF$ due to abscence of cluttering restriction."
Here the content is that $U$ is exact implying $U$ commuting with homology functor.(i.e. Set $L_i(F)$ as left derived functors of $F$. $U(L_iF)\cong L_i(UF)$ was the task of Exercise 2.4.2.)
$\textbf{Q:}$ Such forgetful functor is always exact in $mod-R\to Ab$ case. Why it is "often" exact? Furthermore, computing $L_i(UF)$ is not really computing $L_i(F)$ as certain non-trivial structural results are thrown away if $R\neq Z$. Up to what level of information does $L_i(UF)$ extract?(We surely have abelian group level information but we cannot piece together the information to get back to $mod-R$ for sure.) I even doubt one can use numerical invariants e.g. euler characteristics,... on $L_i(UF)$ to estimate numerical invariants on $L_i(F)$. 
 A: It's hard to state a theorem such as "they're always exact" about forgetful functors because there isn't one global definition of what a forgetful functor is. If it just means faithful then certainly there are some non-exact faithful functors, so it's just that it's often the case (for instance the forgetful functor $mod-R\to Ab$ is always exact)
As for your second question, yes, computing $UL_iF(\simeq L_i UF$) does not provide all the information about $L_iF$ (that's what forgetful means : you lose some information).
But quite often $U$ preserves or reflects what we're interested in : for instance in the example $U:mod-R\to Ab$, $U$ reflects isomorphism, monomorphisms, isomorphisms (if $Uf$ is an isomorphism, so is $f$) and it preserves them, so the information of $UL_iF$ is all we need if what we want is some rough comparison between different things. 
Moreover, it can also happen that there aren't that many possible structures on $UL_iF$, so computing $UL_iF$ may help in identifiying $L_iF$, even if it's just to help for a guess
