The only problem is that you’re looking at the wrong three points: you’re looking at $x+2h,x+h$, and $x$, and the version that you want to prove is using $x+h,x$, and $x-h$. Start with $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\;,$$ and you’ll be fine.
To see that this really is equivalent to looking at $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x+h)-f\,'(x)}h\;,$$ let $k=-h$; then
$$\begin{align*}
f\,''(x)&=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\\
&=\lim_{-k\to0}\frac{f\,'(x)-f\,'(x-(-k))}{-k}\\
&=\lim_{k\to 0}\frac{f\,'(x-(-k))-f\,'(x)}k\\
&=\lim_{k\to 0}\frac{f\,'(x+k)-f\,'(x)}k\;,
\end{align*}$$
and renaming the dummy variable back to $h$ completes the demonstration.
Let $X,Y$ and $Z$ be vector spaces, and $\mathcal{L}(A,B)$ the space of all linear maps from $A$ to $B$.
As noted above, if $F \in \mathcal{L}(X,\mathcal{L}(Y,Z))$, then we can form another map $F_{curry}:X \times Y \to Z$ defined by $F_{curry}(x,y) = F(x)(y)$. Notice that $F_{curry}$ is a bilinear mapping: fixing $x$, $F_{curry}(x,\cdot)$ is linear in the second slot, and $F_{curry} (\cdot,y)$ is linear in the first slot. Conversely, given a bilinear mapping from $G: X \times Y \to Z$, I can produce an element $G_{uncurry}\mathcal{L}(X,\mathcal{L}(Y,Z))$ in the way you would expect: $G_{uncurry}(x)(y) = G(x,y)$. Keyword here is "Curry-howard isomorphism".
So $\mathcal{L}(X,\mathcal{L}(Y,Z))$ can be canonically identified with the space of bilinear mappings from $X \times Y \to Z$. These in tern could be identified with linear mappings from the space $X \otimes Y \to Z$, the so called "tensor product" of $X$ and $Y$, but I will not go into that.
You might be curious how you could work with such an object. What data do you need to write down? For a linear map, you only have to specify action on a basis, but a bilinear map is not a linear map. It turns out (you should check) that specifying the action on all pairs of basis vectors is enough.
Lets get back down to earth and examine a very special case. Let $f:\mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x,y) = x^2y$.
$D(f)\big|_{(x,y)}$ is the linear map given by the matrix $\left[ \begin{matrix} 2xy&x^2\end{matrix} \right]$. That is to say, $D(f)\big|_{(x,y)}(\Delta x,\Delta y) = 2xy\Delta x + x^2\Delta y \approx f(x+\Delta x,y+\Delta y) - f(x,y)$. Notice that the transpose of this matrix is the "gradient" of $f$. Only functions from $\mathbb{R^n} \to \mathbb{R}$ have gradients.
The second derivative should now tell you how much the derivative changes from point to point. If we increment $(x,y)$ by a little bit to $(x+\Delta x,y)$ then we should expect the derivative to increase by about $\left[ \begin{matrix} 2y\Delta x&2x \Delta x\end{matrix} \right]$. Similarly, when we increase $y$ by $\Delta y$, the derivative should change by about $\left[ \begin{matrix} 2x \Delta y&0\Delta y\end{matrix} \right]$.
By linearity, if we change from $(x,y)$ to $(x+\Delta x,y+\Delta y)$, we expect the derivative to change by $$\left[ \begin{matrix} \Delta x&\Delta y\end{matrix} \right] \left[ \begin{matrix} 2y&2x\\2x&0\end{matrix} \right]$$
This gives a matrix which is the approximate change in the derivative. You can then apply this to another vector if you so wish.
Summing it up, if you wanted to see approximately how much the derivative changes from $(x,y)$ to $(x+\Delta x_2,y+\Delta y_2)$ when both are evaluated in the same direction $(\Delta x_1,\Delta y_1)$, you would perform the computation:
$$\left[ \begin{matrix} \Delta x_2&\Delta y_2\end{matrix} \right] \left[ \begin{matrix} 2x&2x\\2x&0\end{matrix} \right] \left[ \begin{matrix} \Delta x_1\\\Delta y_1\end{matrix} \right]$$
The matrix of second partials derived above is called the Hessian, but it a bit misleading to write it as a matrix, since it is really acting as a bilinear form in the manner shown above, i.e. $H(v_1,v_2) = v_1^T H v_2$. You may remember seeing the Hessian arise in multivariable calculus when classifying critical points as maxima, minima, or saddles. In general, the sign eigenvalues of the hessian matrix tell the whole story (although, if there are some zero eigenvalues you might have to climb up the derivative ladder to trilinear forms, etc).
Notice that I only got a Hessian "matrix" because the codomain of $f$ was one dimensional. If it has been, say, $3$ dimensional I would have needed $3$ such matrices, and they would naturally align themselves into a $2\times2\times3$ dimensional box, which would represent a higher order tensor.
Hopefully this gives at least a hint of how to continue. Buzzwords to look for are "multilinear algebra", "tensor products", "tensors", "tensor analysis", and "multivariable taylors theorem".
I do not have a super great reference for this because, even though I do analysis in Several Complex Variables, I have somehow never found a book that treats higher dimensional real analysis really well. I am sure there are books out there, but I have worked out most of this stuff on my own. As far as I know there was never a course offered on it at any university I went to! I guess people are supposed to sort of absorb this stuff when they learn differential geometry.
Best Answer
I might not be exact here, but would you call the relative importance "sensitivity"?
My thought comes from stability analysis, we could think the other way in your case, i.e., how unstable is $r$ with respect to a change in a variable $x$. To do that, I will take $r$ and nudge its input by $\Delta x$ to see how far the new $\tilde{r}:=r+\Delta r$ is:
Comparing $\Delta_s r$, $\Delta_e r$ and $\Delta_c r$, would tell you that the fastest growing one is the most sensitive, i.e., the largest in displacement is the most important to $r$.
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