I did a search on the order of transformations applied to graphs, and mostly found the following, e.g. in this post.
Given a function $f$ always perform transformations $$Af(Bx+C)+D$$ in the order $C,B,A,D$.
But after doing a little digging I'm not sure this is correct. For example, the functions $$y=\frac{1}{\frac{1}{2}x+1}+3\tag{1}$$ and $$y=\frac{2}{x+2}+3\tag{2}$$ are identical.
But suppose $f(x)=1/x$. Using the above approach on (1) transforms $(1,1)$ to $(0,4)$, whilst (2) transforms $(1,1)$ to $(-1,5)$.
Can someone see what might be going on here and perhaps explain? I must have a mental block on this one...
The first function performs the following transformations to $f(x)=\frac{1}{x}$:
while the second performs the following:
As you have noted, these are not the same transformation. However, they both map the graph of $f(x)$ to the graph of $g(x)=\frac{1}{\frac{1}{2}x+1}+3=\frac{2}{x+2}+3$. They do not necessarily map each $(x,y)$ to the same point (in fact you showed they don't), but they both work.
This might seem strange to you, but consider an even simpler example: $$\frac{1}{\frac{1}{2}x}=\frac{2}{x}.$$ The first function says we should stretch $\frac{1}{x}$ horizontally by a factor of $2$, while the second says we should stretch it vertically by a factor of $2$. These are not the same transformations on $\mathbb{R}^2$, but they have the same effect on the function $\frac{1}{x}$.
For another example, consider $f(x)=\ln(x)$. We know that $\ln(ax)=\ln(a)+\ln(x)$. So for this choice of $f(x)$, compressing the function horizontally by a factor of $a$ is equivalent to shifting it vertically by $\ln(a)$ units. Again, these are not the same transformation on $\mathbb{R}^2$, but they have the same effect on the function $\ln(x)$.