= Solution
Take compactly supported <metric variations>, or impose <boundary conditions> that remove the <integration by parts> terms. Let $F=f'(R)$ and $\Box=\nabla^a\nabla_a$. Since $\delta\sqrt{-g}=\frac12\sqrt{-g}\,g^{ab}h_{ab}$, the gravitational action varies as
$$
\delta S_g=\int\sqrt{-g}\left[
\left(\frac12f g^{ab}-F R^{ab}\right)h_{ab}
+F\left(\nabla^a\nabla^b h_{ab}-\Box h\right)\right]d^4x.
$$
Applying <integration by parts> twice to the derivative terms gives
$$
\delta S_g=-\int\sqrt{-g}\,E^{ab}h_{ab}\,d^4x,\qquad
\boxed{E_{ab}=F R_{ab}-\frac12f g_{ab}-\nabla_a\nabla_bF+g_{ab}\Box F.}
$$
For the usual <stress-energy tensor> definition $T_{ab}=-2(-g)^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant <metric tensor> gives $\delta S_m=\frac12\int\sqrt{-g}\,T^{ab}h_{ab}\,d^4x$. Thus the action's stated normalization implies
$$
\boxed{E_{ab}=\frac12T_{ab}.}
$$
There is no implicit $1/(16\pi)$ in this gravitational action. This is the <normalization of the metric f(R) field equation>, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.
Finally, the <chain rule> gives $\nabla_a\nabla_bF=f''\nabla_a\nabla_bR+f^{(3)}\nabla_aR\nabla_bR$ and $\Box F=f''\Box R+f^{(3)}(\nabla R)^2$. Substitution yields
$$
\boxed{E_{ab}=f'R_{ab}-f''\nabla_a\nabla_bR-f^{(3)}\nabla_aR\nabla_bR
+\left(-\frac12f+f''\Box R+f^{(3)}\nabla_cR\nabla^cR\right)g_{ab}.}
$$
This is a metric <f(R) gravity> variation: the connection is always the Levi-Civita connection of the varied <metric tensor>, not an independent variable.
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