What does $\liminf_{x\to0+}-\frac{f'(x)}x>0$ informally mean? Let $R>0$ and $f:(-R,R)\to[0,\infty)$ be nonincreasing on $[0,R)$ with $$f(x)=f(-x)\;\;\;\text{for all }x\in(-R,R)\tag0.$$

What does $$\alpha:=\liminf_{x\to0+}-\frac{f'(x)}x>0\tag1$$ informally mean?

From $(1)$ we see that, with $c:=\alpha/2>0$, there is a $R_0\in(0,R)$ with $$f'(x)<-cx\;\;\;\text{for all }x\in(0,R_0]\tag2.$$

So, is the meaning of $(1)$ that the "rate of decay of $f$ is at least linear in a neighborhood of $0$" or something like that?

 A: Probably, the simplest thing is to focus on the derivative of $f$. Since $f$ is an even function, its derivative $f'$ will be odd:
$$f'(x)=-f'(-x)$$
Moreover, in your formula (1), one can note that since $f$ is even in $(-R,R)$, its derivative in zero, will be:
$$f'(0):=\lim_{h\to 0^+} \frac{f(h)-f(-h)}{2h} = 0$$
Also, by the fact that $f$ is non-increasing in $[0,R)$ we recover that $f'(x)\le 0$ forall $x\in [0,R)$.
At this point,
$$\alpha = \liminf_{x\to 0^+}-\frac{f'(x)}{x}=-\limsup_{x\to 0^+}\frac{f'(x)-f'(0)}{x}$$
can be interpreted as an upper bound for the second derivative (if well defined) in a neightbourhood of $0$, which, as you said means that
$$f'(x)\le -\alpha x$$
for $x$ sufficiently small. At this point, going back to $f$, we can interpret this condition as the fact that
$$f(x)\le f(0)-\frac{\alpha}{2} x^2$$
in a sufficiently small range around $0$.
Formally, is should be
$$\forall \eta<\alpha\ \exists R_0>0\qquad  f(x)\le f(0)-\frac{\eta}{2} x^2\ \ \forall x\in [-R_0,R_0]$$
but I think that If you are looking at an interpretation, the first formulation is much better
