Consider the language in the following example:
What is meant by $dg$ and $dR$, and also by $dg/g$? Why does $dR/R=-2/100$ (negative for shrinks)? Is $4\%$ unity change? I mean $dg/g=4\%$ or $dg=4\%$
I'm primarily interested in the general case of notation of the form $dy/y$ for arbitrary $y$.
In the example $dR$ means "change in $R$" and $dg$ means "change in $g$". In particular in this case you start with a radius $R$, which then changes to some other value $R'=R+dR$. This change thus causes a change in $g$, which becomes $g'=g+dg$.
So now, if $R$ shrinks by $2\%$, this means that $R' = R-0.02\times R$, which in terms of the change gives $dR = -0.02R$ or equivalently $dR/R=-0.02=2/100$.
To answer your other question, it is $dg/g=4\%$. This follows from the algebra that they do in the example. Notice that it can't be $dg=4\%$ because $4\%=0.04$ is just a number with no dimension, while $dg$ being a change in $g$ must have the same units of $g$, that is those of an acceleration.
In general, whenever you find an expression of the form $dy/y$ that is comparing the change in $y$ to $y$ itself. This notation with differentials is meant to emphasise the fact that the changes are small compared to the quantities involved (that is for example $dR$ is much smaller than $R$, because it is only $2\%$ of it). For this reason they're using calculus to get to the result.
Suppose you have that dg/dR = -2g/R. The entire expression dg/dR is an atomic symbol representing the derivative of g with respect to R. While it is presented as a fraction, it is not technically a fraction in the normal sense. When you have d(something) by itself, this is known as an "infinitesimal". Infinitesimals aren't like "normal" numbers, but generally speaking when you divide one by another, their weirdness "cancels out", and we can treat the ratio as being a normal number. So even though infinitesimals aren't normal numbers, for a lot of purposes, we can pretend they're normal numbers, and multiply both side by the one of them. In this case, if we multiply both sides by dR, that gives us dg = -2 dR/R.
In the example you give, rather than starting with dg/dR = -2g/R, the example starts with log(g) = log(G)+log(M)-2log(R), and takes the derivative of both sides. Normally, when you take the derivative, you take the derivative of one thing with respect to a second thing, and you say that d(first thing)/d(second thing) = derivative. However, there's a concept of an "implicit" derivative where you take the derivative of each term, but instead of taking the derivative with respect to something, you just leave the infinitesimal for the variable in that term. So when you have log(g), the derivative of that is 1/g, and then you include the infinitesimal for g, getting dg/g. Similarly, we get dR/R on the other side (G and M can be ignored because the are constants, and thus their derivatives are zero).
If we want the explicit derivative, we'll have to divide both sides by dR. However, when you have dR/R, notice that you're dividing the change in R by the total R. That can be thought of as a percent (a percent is part of something divided by the total). Since we're given a percent, and we're asked for another percent, it's easier to just keep it in the form dR/R.
This example is a bit imprecise, however; it treats the instantaneous rate of change (derivative) as being the same as the overall rate of change. For small percentages, this is a decent approximation, but it wouldn't work for larger ones. Treating dR/R, or dy/y, as being "equal" to something is playing a bit loose with calculus. Strictly speaking, you have to have a ratio between two such expressions. For instance, if dy/y = 2dx/x, then that means that if x increases by a certain percentage, then y will increase by about twice that percentage (and it will get closer to exactly twice the closer the percentage is to zero). If y increases by 4%, then x must have increased by about 2%.
Also, in the example, there's a bit of a space between d and R, which is probably a typesetting issue. It should be -2dR/R, not -2d R/R.
The other answers gave some useful insights. I just want to look a little more at the differentiation of $\log g$:
Normally, when we differentiate a function $f(x)$ with respect to the variable $x$, we might write
$$\frac{df}{dx} = f'(x)$$
In the case when the function is $\log(x)$, we would find
$$\frac{df}{dx} = \frac{d(\log(x))}{dx} = \frac{1}{x}$$
We can rearrange this to give
$$df = \frac{dx}{x}$$
In other words - a small change in the function $\log(x)$ when $x$ changes by a small amount $dx$ depends on the value of $dx$ and $x$ : the larger $x$ is, the smaller the change in $\log(x)$ for a given change in $x$.
This kind of technique can be useful to see "how much does something change when something else changes by a certain amount" (which is of course the subject of calculus). In the case of your problem, having first written the expression for the acceleration of gravity
$$g = \frac{GM}{r^2}$$
we can now take the logarithm on both sides, and then differentiate:
$$\log(g) = \log(G) + \log(M) - 2\log(r)$$
And now we can look at the fractional change in each of these things when another thing changes. Since $G$ and $M$ are constant in this problem, their derivative is zero. We can look at what happens to the other two values by using the trick I showed above, and we get
$$\frac{dg}{g} = -2\frac{dr}{r}$$
In other words, the fractional change in $g$ is twice as big as the fractional change in $r$: if the radius changes by -2% (getting smaller), the acceleration of gravity changes by 4% (getting bigger)
