Why are horizontal transformations of functions reversed?
You're really talking about what happens to the graph $y=f(x)$; and from this perspective, we can see that horizontal (x) and vertical (y) transformations work the same way.
Instead of writing $y=f(x)+a$, write $y+b=f(x)$ (here, $b=-a$); and instead of $y=af(x)$, write $by=f(x)$ (here, $b=\frac{1}{a}$).
So for translations, we have
- $y=f(x+a)$ shifts $x$ by $-a$.
- $y+b=f(x)$ shifts $y$ by $-b$.
And for scaling, we have
- $y=f(ax)$ scales $x$ by $1/a$.
- $by=f(x)$ scales $y$ by $1/b$.
So you see, they really work the same way, it just looks opposite because the factor $a$ gets moved to the other side.
This works very generally. Suppose we have
- equation 1: $F(x,y,z)=0$,
- equation 2: $F(x,y,z+c)=0$, and
- equation 3: $F(x,y,dz)=0$.
Now take any solution to equation 1, lets call it the triple $(n_1,n_2,n_3)$. (So equation 1 is true if I plug in the numbers $n_1$ for $x$, $n_2$ for $y$, and $n_3$ for $z$.)
Then you can see that $(n_1,n_2,n_3-c)$ is a solution of equation 2, and $(n_1,n_2,\frac{1}{d}n_3)$ is a solution of equation 3.
Jonas Kibelbek's answer covers almost all of what I'd have said. The one thing I'd add is that the substitution $$x\mapsto\frac{x-h}{a}$$ (or similarly $y\mapsto\frac{y-k}{b}$) is a dilation by a factor of $a$ centered at $0$ (if $|a|>1$, it's a stretch; if $|a|<1$, it's a shrink), followed by a translation by $h$ (if $h>0$, in the positive direction (and similarly for $y$, $b$, and $k$). One way to think of this is to change the way we're writing the mapping a bit (still talking about the same mapping, just writing it differently): $$\begin{align} x&\mapsto\frac{x-h}{a} \\ x_{\text{old}}&=\frac{x_{\text{new}}-h}{a} \\ ax_{\text{old}}&=x_{\text{new}}-h \\ ax_{\text{old}}+h&=x_{\text{new}} \end{align}$$