This is another grey area with mathematical terminology. Inversion can mean many different things in many different instances, but in general, it has to do with swapping two things, or creating something's "opposite." Each definition you give of inversion or finding an inverse can be made more precise with more words: $1/a$ is the multiplicative inverse of $a$, $-a$ is the additive inverse of $a$, $f^{-1}(x)$ is the inverse function of $f(x)$, etc. It can become confusing if one is not careful in context. For instance, suppose we are just talking about functions, and I bring up the topic of inverses. It is clear from context that every inverse we talk about is going to be the inverse function. If we want to make reference to $1/a$ or $-a$ at that time, we might instead use the words reciprocal and opposite, respectively. It just happens that saying "inverse" instead of the more descriptive "multiplicative inverse" say, is more brief and can at times be unnecessary.
All in all, there can be many different terms for a single operation, and there can be one term that applies to many different operations. This is quite common in introductory mathematics, as the different words are used to convey different meanings within context. You can see this with terms like "inversion" and "prime," but in each setting, you should strive to make their meanings precise and leave no ambiguity (but certainly within reason).
(P.S. Yes, arcsine is the same as inverse sine. Here is the functional inverse)
Perhaps not quite the way you are looking for, but:
You can derive Taylor's theorem with the integral form of the remainder by repeated integration by parts:
$$ f(x)-f(a) = \int_a^x f'(t) \, dt = \left[-(x-t)f'(t) \right]_a^x + \int_a^x (x-t) f''(t) \, dt \\
= (x-a)f'(a) + \int_a^x (x-t) f''(t) \, dt, $$
and so on, integrating the $(x-t)$ and differentiating the $f$ each time, to arrive at
$$ R_N = \int_a^x \frac{(x-t)^N}{N!} f^{(N+1)}(t) \, dt. \tag{1} $$
Interpretation for this is simply that integrating by parts in the other direction will give you back precisely $f(x) - f(a) - \dotsb - \frac{1}{N!}(x-a)^N f^{(N)}(a)$.
Now, we can get from (1) to the Lagrange and Cauchy forms of the remainder by using the Mean Value Theorem for Integrals, in the form:
Let $g,h$ be continuous, and $g>0$ on $(a,b)$. Then $\exists c \in (a,b)$ such that
$$ \int_a^b h(t) g(t) \, dt = h(c) \int_a^b g(t) \, dt. $$
(this is easy if you think about weighted averages and the usual Mean Value Theorem).
Applying this to (1) with $h=f$, $g(t)=(x-t)^N/N!$ gives
$$ R_N = f^{(N+1)}(c)\frac{(x-a)^{N+1}}{(N+1)!}, $$
which is the Lagrange form of the remainder; using $h(t)=f(t)(x-t)^N/N!$, $g(t)=1$ gives
$$ R_N = f^{(N+1)}(c')\frac{(x-c')^N}{N!}(x-a), $$
which is the Cauchy form of the remainder.
The weighted averages mentioned above are a way to think about what we did here: we take an average of $\frac{(x-t)^N}{N!} f^{(N+1)}(t)$ over $[a,x]$, and use the MVT to equate this to a value of the function at a specific point: how much of the function we count as in the weighting affects what answer we obtain.
Best Answer
To say "$f$ increases in proportion to $g$" is just another way of saying "$f$ is proportional to $g$", i.e., $f$ is equal to $g$ times a nonzero constant.
(In other contexts this may only be meant approximately, but the "literal" usage is that it is equal to $g$ times a constant.)