Let $f$ be a scalar field defined on a set $S$ in $\mathbb{R}^n$ and let $\pmb{a}$ be an interior point of $S$.
When we study how this field changes as we move from $\pmb{a}$ to a nearby point, we use the quotient
$$\frac{f(\pmb{a}+h\pmb{y})-f(\pmb{a})}{h}\tag{8.3}$$
As Apostol puts it
The numerator of this quotient tells us how much the function changes when we move from $\pmb{a}$ to $\pmb{a}+h\pmb{y}$. The quotient itself is called the average rate of change of $f$ over the line segment joining $\pmb{a}$ to $\pmb{a}+h\pmb{y}$. We are interested in the behavior of this quotient as $h\to 0$.
Now assume $\pmb{y}$ is a unit vector.
The distance between $\pmb{a}$ and $\pmb{a}+h\pmb{y}$ is now $|h|$.
In this case the difference quotient (8.3) represents the average rate of change of $f$ per unit distance along the segment joining $\pmb{a}$ to $\pmb{a}+h\pmb{y}$.
My question is about this subtle change in interpretation of 8.3 when we make $\pmb{y}$ a unit vector.
Let me try to interpret what happens when we have a real-valued function of one variable.
I'm going to use vectors here just so that I can keep the parallel sort of thinking with the higher dimension case.
Let $\pmb{y}=2\hat{i}$ and $\pmb{a}=a\hat{i}$. Then,
$$\pmb{a}+h\pmb{y}=(a+2h)\hat{i}$$
and the quotient is
$$\frac{f(a+2h)-f(a)}{h}\tag{1}$$
On the other hand, if $\pmb{y}=\hat{i}$ then it is a unit vector and the quotient is
$$\frac{f(a+h)-f(a)}{h}\tag{2}$$
If we take limits as $h\to 0$ of (1) and (2) we get, according to the theory in Apostol's Calculus, the derivative of the scalar field $f$ with respect to the vector $\pmb{y}=2\hat{i}$ at $\pmb{a}$, and the derivative of $f$ with respect to $x$ at $\pmb{a}$ , respectively. The latter is also the directional derivative of $f$ at $\pmb{a}$ in the direction of $\pmb{y}=\hat{i}$.
My question is, what is an intuitive way to understand the difference between
$$\lim\limits_{h\to 0}\frac{f(a+2h)-f(a)}{h}\tag{3}$$
and
$$\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}\tag{4}$$
in this one-dimensional case, but also in the n-dimensional case?
In particular, I don't really understand the phrase "average rate of change of $f$ over the line segment joining $\pmb{a}$ to $\pmb{a}+h\pmb{y}$".
For example, let $f(\pmb{x})=\lVert \pmb{x} \rVert^2$ and $\pmb{a}=\hat{i}$.
Then,
$$f(\pmb{a})=1$$
$$f(\pmb{a}+2h\hat{i})=(1+2h)^2$$
When $h=1$ then
$$\frac{f(\pmb{a}+h\pmb{y})-f(\pmb{a})}{h}=8$$
What this means is that the average rate of change of $f$ over the line segment between 1 and 3 is 8. It's as if the "unit" is 2, ie the "unit" step we take is the length of the vector the direction of which we are taking the derivative in.
After writing this all out, it seems to me that "average rate of change over the line segment" means "average rate of change per length of $\pmb{y}$ in the direction of $\pmb{y}$", and in the special case of a unit vector $\pmb{y}$ we have "average rate of change per unit distance in the direction of $\pmb{y}$.
Given a function $f$ in $\mathbb R^n$, the differential $df_p$ in a point $p\in \mathbb R^n$ is a linear map sending vectors leaving $p$ to real numbers $\mathbb R$. It is defined in this way: $$ df_p(v)=(f\circ \alpha)'(0) $$ where $\alpha(t)$ is any curve such that $\alpha(0)=p$ and $\alpha'(0)=v$, and $'$ means the usual definition of derivative. It is, in some sense, the best linear map approximating $f$ near $p$, and the meaning is the "average variation" of $f$ between $p$ and $p+v$.
In the particular case of a real function, the usual derivative is a particular case of this: $$ df_a(1)=(f\circ \alpha)'(0) $$ with $\alpha(t)=a+t$, and then $$ df_a(1)=(f\circ \alpha)'(0)=\lim_{t\to 0}\frac{f(a+t)-f(a)}{t} $$
And now, what is $df_a(2)$? Consider the curve $\beta(t)=a+2t$ and then $$ df_a(2)=(f\circ \beta)'(0)=\lim_{t\to 0}\frac{f(a+2t)-f(a)}{t} $$
So the answer is that both expression are related because they are the same linear map applied to different vectors.