I am currently writing an expository paper on gauge theory and gravity and throughout the gauge theory part I have been able to mostly stay away from coordinates unless I wished to provide specific examples. I am trying to continue this practice through the section on general relativity but am struggling at writing a coordinate free proof of the variation of the Einstein-Hilbert action.

Ideally what I am looking for is a way to take the Einstein-Hilbert action : $$ S[g]=\int_M\langle g,Rc\rangle_g\text{dvol}_g $$ (where $\langle\cdot,\cdot\rangle$ is the induced inner product on the bundle of symmetric tensors, and $\text{dvol}_g$ is the volume form) and then vary it with a section of the symmetric tensor bundle $h$: $$ \frac{d}{dt}\Big|_{t=0}S[g+th]=0 $$

A similar question was asked here many years ago, and I have tried to follow up on their recommended sources, namely Besse's Einstein Manifolds and I am honestly struggling to figure out how they are obtaining the results they are obtaining as they do not provide many details.

For concrete examples of where I am struggling, take the variation of the volume form. Besse states that: $$ \begin{align} \frac{d}{dt}\Big|_{t=0}\text{dvol}_{g+th}=\frac{1}{2}\text{tr}_g(h)\text{dvol}_g \end{align} $$ and I agree, but I can only get there in coordinates by writing: $$ \begin{align} \frac{d}{dt}\Big|_{t=0}\text{dvol}_{g+th}=&\frac{d}{dt}\Big|_{t=0} \sqrt{\det(g+th)}dx^1\wedge \cdots \wedge dx^n\\ =& \frac{1}{2}\frac{1}{\sqrt{\det(g)}}\frac{d}{dt}\Big|_{t=0}\det(g+th)dx^1\wedge \cdots dx^n\\ =&\frac{1}{2}\frac{1}{\sqrt{\det(g)}}\det(g)\text{tr}(g^{-1}h)dx^1\wedge \cdots \wedge dx^n\\ =&\frac{1}{2}\sqrt{\det(g)}g^{ij}h_{ij}dx^1\wedge \cdots \wedge dx^n\\ =&\frac{1}{2}\text{tr}_g(h)\text{dvol}_g \end{align} $$

Furthermore, when varying the scalar curvature, $s=\langle g,Rc\rangle_g$ we writes that this most easily found by noting that $s=\text{tr}_g(Rc)$ and then applying the product rule: $$ \begin{align} \frac{d}{dt}\Big|_{t=0}\text{tr}_{g+th}(Rc)=\langle h , Rc\rangle _g+\text{tr}_g\left(\frac{d}{dt}\Big|_{t=0}Rc_{g+th}\right) \end{align} $$ I can see how the second term comes about as the trace of $Rc$ with respect to $g$ is just $g^{-1}\lrcorner Rc$, but the first term is harder for me to come by without coordinates. With coordinates I can write that: $$ \begin{align} \frac{d}{dt}\Big|_{t=0}\text{tr}_{g+th}(Rc)=&\frac{d}{dt}\Big|_{t=0}(g+th)^{ij}Rc_ij\\ =&-g^{ik}h_{kl}g^{jl}Rc_{ij}+g^{ij}\frac{d}{dt}\Big|_{t=0}Rc_ij\\ =&\langle h , Rc\rangle _g+\text{tr}_g\left(\frac{d}{dt}\Big|_{t=0}Rc_{g+th}\right) \end{align} $$ But is there any way to see this without coordinates? Perhaps some notation that allows one to write the inner product in terms of $g$ without reference to coordinates? I feel like if I can past my initial uneasiness here I'll be able to follow the rest of the argument in the book, albeit with some work, so any help would be appreciated.