Many students start calculus by learning to integrate over intervals on the real line. Then they integrate functions over 2D and 3D domains, adding more integral signs to match the number of dimensions. I suppose this is nice notation in the typical cases where you can do your integral dimension-by-dimension, e.g.
$$ \iiint\limits_V f(x, y, z) \,\mathrm{d}V = \int_{x_0}^{x_1} \left(\int_{y_0(x)}^{y_1(x)} \left(\int_{z_0(x,y)}^{z_1(x,y)} f(x, y, z) \,\mathrm{d}z \right) \,\mathrm{d}y \right) \,\mathrm{d}x. $$
The three integrals on the left remind you that you can reduce this multidimensional problem to nested single-dimensional problems. If you are really lucky, the whole thing separates and you have
$$ \iiint\limits_V f(x) g(y) h(z) \,\mathrm{d}V = \left(\int_{x_0}^{x_1} f(x) \,\mathrm{d}x\right) \left(\int_{y_0}^{y_1} g(y) \,\mathrm{d}y\right) \left(\int_{z_0}^{z_1} h(z) \,\mathrm{d}z\right). $$
But that triple integral was just a single symbol. What did it mean? Basically two things: (1) sum up the contributions over some domain of some integrand to follow, and (2) this domain is a subset of $\mathbb{R}^3$. Now point (2) is already conveyed in the other symbol $\mathrm{d}V$, which can be written $\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ or $r^2 \sin(\theta) \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\phi$ or any number of other ways. In fact, these expanded forms are even more informative, conveying some information about the interpretation of the variables. As a result, all we really need the integrals for is to convey point (1), and a single symbol does this just as well as three.
For this reason, it is just as correct to write
$$ \int\limits_V f \,\mathrm{d}V $$
as it is to write
$$ \iiint\limits_V f \,\mathrm{d}V. $$
They are perfectly equivalent.
Having this mindset is also more useful in more general/advanced branches of analysis, where integration is closely tied to measure theory, and one can put measures on $\mathbb{R}^3$ just as easily as on $\mathbb{R}$, so there isn't much point in distinguishing.
In summary, feel free to use whichever notation you like, though it's usually a good idea to match the notation of your texts/instructors, at least until you are familiar enough to decide for yourself which is most reasonable.
Here is verification that the function satisfies the DE. Uniqueness depends on the function $f$ being "nice enough".
When you differentiate the integral, you have to treat it like a product since the variable you are differentiating w.r.t. ($t$), shows up in both the integrand and limits.
Best Answer
Let's work backwards from the end of your question:
No, it's not. The latter expression is (almost, but not quite) the derivative of the former expression.
This might be easier to see if you introduce a change of variable to reduce some of the noise in the formula. Let's write $u(t) = \frac{dR}{dt}$. Then the first expression is $\frac12 \left( u(t) \right)^2$. It's derivative (with respect to $t$) is, by the Chain Rule, $u(t) \cdot u'(t)$. That is, $$\frac{d}{dt} \left( \frac12 \left(\frac{dR}{dt}\right)^2 \right) = \frac{dR}{dt}\cdot \frac{d^2R}{dt^2}$$
Now let's try to understand what the text is saying.
I suspect there is a typo in here, as it should say "Multiplying both sides by $dR_s/dt$". If we do this, the original equation becomes $$\frac{d^2R_s}{dt^2} \cdot \frac{dR_s}{dt} =-\frac{GM_s}{R_s(t)^2} \cdot \frac{dR_s}{dt}$$
Now let's integrate both sides with respect to $t$. We have $$\int \frac{d^2R_s}{dt^2} \cdot \frac{dR_s}{dt} \, dt =- \int \frac{GM_s}{R_s(t)^2} \cdot \frac{dR_s}{dt} \, dt$$
Let's tackle these two integrals separately. For the left-hand side, we will use the substitution $u(t) = \frac{dR}{dt}$. (This is the same substitution I used at the start of this answer!) Then the left-hand side reads $\int u'(t) \cdot u(t) \, dt$. This is exactly the same thing as $\int u \, du$, which integrates easily to $\frac12 u^2$. In other words, the left-hand side is $\frac{1}{2}\left(\frac{dR}{dt}\right)^2$.
Now for the right-hand side. This time set $v = R_s(t)$. Then (setting aside some constants) the integral is $\int v^{-2} dv$, which is easy to integrate, and we get $-\frac1v$. So the integral on the right is $\frac{GM_s}{R_s(t)}$.