I edited the whole question, even the title. Originally, I wanted to know if non-linear functions could have the following property: Same amount of change in $x$ always corresponds to a constant change in $f(x)$.
From Cauchy's functional equations, under regularity conditions, it follows that if $f(x+\Delta x) = f(x) + f(\Delta x)$, then $f$ is a linear function. But it doesn't answer my question completely, since $f(\Delta x)$ is a certain type of constant.
I am assuming that this is meant to be a function from real numbers to real numbers.
It follows by induction from what you write that
$$g(n\Delta x)=mn\Delta x+c$$
for all integers $n$, where $c=g(0)$. If the property is meant to hold for all possible real number shifts $\Delta x$*, then we can choose $\Delta x=1/l$ for an integer $l$ to get
$$g(n/l)=m(n/l)+c$$
So, for rational inputs ($x$ is rational if it is the ratio of two integers), the only possible functions with this property are in the form $g(x)=mx+c$.
You may be interested in reading
https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation
Your equation is an example of a functional equation, related to the one in the article.
*If the condition holds for a particular value of $\Delta x$ only, then there are clearly more solutions. Namely, if we add to $g$ any periodic function with period $\Delta x$.
*To answer the new, edited question, if $f(x+\Delta x)-f(x)=g(\Delta x)$ for all $x,\Delta x$ then, for rationals $x$, $f(x)=mx+c$ for some real $m,c$ and $g(x)=mx$ for the same $m$. Again, there are solutions which don't take this form. To describe them, we would use a $\mathbb{Q}$-basis for $\mathbb{R}$, i.e. a set $B$ of real numbers such that any real number may be written uniquely as a finite sum $\sum a_ib_i$ with $b_i\in B$ and $a_i\in \mathbb{Q}$. Given such a set we define $f(\sum a_ib_i)=\sum a_im_i+c_i$ for reals $m_i,c_i$ and $g(\sum a_ib_i)=\sum a_im_i$.
Note that your function is not quite linear but affine. since $g(x+y)$ is not equal to $g(x)+g(y)$. But I know what you mean.
So suppose there exists a function $g$ such that $g(x+h)-g(x)=mh$ for all $h$ and $x$.
then dividing by $h$ and taking the limit as $h$ tends to $0$ we see:
$$g'(x)=m$$
Integrating the above equation we get $g(x)=mx+a$ for some constant $a$.
Edit:
To show rigorously that the limit exists, note that
$\dfrac{f(x+h)-f(x)}{h} - m = \dfrac{mh}{h} - m = 0$
For all $h$, hence the limit tends to zero. As you can see, this was quite trivial to show.
Suppose that $f(x+\Delta x)=f(x)+g(\Delta x)$.
If $0$ is in the domain of $f$ then $f(0+\Delta x)=f(0)+g(\Delta x)$.
Thus $f(\Delta x)=f(0)+g(\Delta x)$ for $\Delta x$ in the domain of $f$.
In other words.
$$ g(x)=f(x)-f(0) $$
This places constraints on the function $f$. For example, there is no such function $g$ for a quadratic function, as is easily demonstrated. However, as others have pointed out, such a function $g$ exists for affine functions.
If $f(x)=ax+b$, then $g(x)=f(x)-f(0)=ax$.
So $f(x+\Delta x)=a(x+\Delta)x+b=ax+b+a\Delta x=f(x)+g(\Delta x)$
Unless $b=0$, $f(x)=ax+b$ is not linear since $f(x+\Delta x)\ne f(x)+f(\Delta x)$.
