Composing Translations and Reflections Since $f(-x)$ is a reflection of $f(x)$ in the $y$-axis and $f(x+a)$ is a shift of $f(x)$ by $-a$ units, so for the longest time I've assumed $f(-(x+a))$ is a shift of $-a$ and then reflected in the $y$-axis.
But graphing some things, I've noticed this actually gives a shift of $+a$, then reflection in $y$-axis? Rewriting $f(-(x+a))$ as $f(-x-a)$ reads reflect, then shift by $+a$ to me, which is equivalent to the former, but produces the latter. 
Am I reading the order of operations wrong here?
 A: This is always confusing (to me as well), and I find myself having to write it down every time -- and even then, a mistake is easily made.
Let us we write $S$ for the reflection operation $f(x) \mapsto f(-x)$ and $T_a$ for the "translation by $a$"-operation $f(x) \mapsto f(x-a)$.
The crux here is that $S$ is not the same as the function $x \mapsto -x$. Similarly, $T_a$ is not the same as $x \mapsto x-a$.

Let us see what is going on. We must consider the following compositions (I suppress the composition symbol $\circ$ for readability):
$$ST_a,\  ST_{-a}, \ T_aS,\ T_{-a}S$$
So, let's take an arbitrary function $f$, and apply the definitions:
\begin{align}
T_a (f) &= x \mapsto f(x-a) \\
T_{-a}(f) &= x \mapsto f(x+a) \\
S(f) &= x \mapsto f(-x)
\end{align}
What is important to keep in mind, is that $ST_a(f) = S(T_a(f))$. So $S$ is applied to the new function $x \mapsto f(x-a)$, and not to the entire argument of $f$, that is, to $x-a$. So, in full detail, we obtain:
\begin{align}
ST_a(f) &= S(x \mapsto f(x-a)) = x \mapsto f((-x)-a) = f(-x-a)\\
ST_{-a}(f) &= S(x \mapsto f(x+a)) = x\mapsto f((-x)+a) = f(a-x)\\
T_aS(f) &= T_a(x \mapsto f(-x)) = x \mapsto f(-(x-a)) = f(a-x)\\
T_{-a}S(f) &= T_{-a}(x \mapsto f(-x)) = x \mapsto f(-(x+a)) = f(-x-a)
\end{align}
(I hope that the $x \mapsto \dots$ notation is not too confusing for you.)
In conclusion, we have that $ST_a = T_{-a}S$ and $ST_{-a} = T_aS$. Intuitively:

"If you pull something towards you from a mirror, then looking in the mirror, you see its mirror image moving away from you."


Added (as suggested by Trevor Wilson):
What the above amounts to, is that we can write $f(-(x+a))$ as $(ST_a(f))(x)$ or as $(T_{-a}S(f))(x)$ (because, and this is important, $ST_a(f)$ and $T_{-a}S(f)$ are still functions, which we can evaluate at $x$). In more advanced treatises, the extraneous brackets are omitted, to yield: $$f(-(x+a)) = ST_af(x) = T_{-a}Sf(x)$$

Further addition -- generalities on transformation operators:
So far, we've been kind of vague on what a transformation operator exactly is, and dealt with examples.
Now, suppose that we have a mapping $g: \Bbb R \to \Bbb R$ (usually, but not necessarily, a bijection). E.g. $g_1(x) = -x$, or $g_2(x) = x-a$.
Then the operator defined by $g$ (which I will denote $O(g)$) is the mapping $O(g): \Bbb R^{\Bbb R} \to \Bbb R^{\Bbb R}$ which takes a function $f:\Bbb R \to \Bbb R$ to: $$O(g)f: \Bbb R\to\Bbb R, x \mapsto f(g(x))$$
We sometimes refer to $O(g)$ as precomposition by $g$: it maps $f$ to $f \circ g$: "$g$ before $f$".
Recalling the examples $g_1$ and $g_2$, we see $S = O(g_1)$ and $T_a = O(g_2)$.
Now, let us see what happens if we look at $O(gh)$ (again, I suppress $\circ$ for readability) for some suitable functions $g,h$:
$$O(gh)f = fgh = O(h)(fg) = O(h)O(g)f$$
Ah-hah! $O$ reverses the order of composition. 
Now let us return to our concrete setting. For conceptual clarity, let us write $t_a$ for the mapping $x \mapsto x-a$ (so that $T_a = O(t_a)$), $s$ for $x \mapsto -x$ and $m_b$ for $x \mapsto bx$.
If we would want to write $f(bx+a)$ using translation and scaling, we'd observe that:
$$bx+a = t_{-a}(bx) = t_{-a}m_b(x)$$
Switching to the $O$ notation, we get:
$$f(bx+a) = O(t_{-a}m_b)f(x) = O(m_b)O(t_{-a})f(x) = M_bT_{-a}f(x)$$
In conclusion, the fact that $O$, precomposition, is composition-reversing gives rise to the counterintuitive order of operations. I hope this provides you with a better understanding of what's going on.
A: I think you are misusing the following rule:

(1) The graph of $f(-x)$ as a function of $x$ is obtained by reflecting the graph of $f(x)$ as a function of $x$ in the vertical axis.

You appear to be using something like the following statement, which is not vaild.

(2) The graph of $f(-(x+a))$ as a function of $x$ is obtained by reflecting the graph of $f(x+a)$ as a function of $x$ in the vertical axis.

The problem is confusing because statement (2) happens to be "almost" correct in the sense that it only differs from the correct statement by a translation.
So to get a sense of why we should not expect statements along the lines of (2) to be valid in general, let's replace the function $x \mapsto x+a$ in (2) by a totally different function, say $x \mapsto x^2$.  Then the same reasoning that gave us (2) would give us

(3) The graph of $f(-x^2)$ as a function of $x$ is obtained by reflecting the graph of $f(x^2)$ as a function of $x$ in the vertical axis.

But statement (3) is clearly false: The graph of $f(-x^2)$ as a function of $x$ depends only on the values of $f$ on arguments $\le 0$ and may be wildly different from the graph of $f(x^2)$ as a function of $x$, which depends only on the values of $f$ on arguments $\ge 0$.
