What are boundary conditions?
Given a recurrence/function $f$ boundary conditions are simply a few points, usually small points such as 0,1 or 2, at which $f$ is specified. For example, in the Fibonacci sequence $$F(n) = F(n-1) + F(n-2), F(0) = F(1) = 1$$
$F(0),F(1)$ are boundary conditions. In other words initial conditions, base conditions.
The excerpt you posted proves the upper bound for the recurrence relation $2T(\lfloor n/2 \rfloor) + n$. It is done using substitution method for solving recurrence relation where you first guess the solution (involving constant(s)) and then find constant(s) that would satisfy boundary conditions.
After you guessed the solution you get an explicit (as in your example) equality/inequality of the form $T(n) \leq cg(n)$ which you have to prove in order to establish the upper bound for the recurrence relation $T(n)$. That is usually done using induction. However, the original boundary conditions may fail when you use it with your guessed function $g(n)$ like in the example: $T(1) = 1$, but $c\log(1) = 0$. At this point you should kind of select new boundary conditions (initial points) to prove the inequality $T(n) \leq cg(n)$. Without base step you cannot complete induction proof.
So the example "overcomes this obstacle" by choosing a new value $n_0$ for the inductive proof. However, $T(1) = 1$ is still kept but it is removed from the inductive proof. The example then introduces new boundary conditions $T(2)$ and $T(3)$ instead of $T(1)$. Given $T(1)= 1$, we have $T(2)= 4$ and $T(3)=5$ and so for some constant $c \geq 2$ the following conditions hold $T(2) \leq c2\log(2)$, $T(3) \leq c3\log(3)$.
Now you can prove that $T(n) < cn\log(n)$ using induction by taking base step $n=2$ and $n=3$ ($n \geq 2$).
How do we know the base case?
If base case (boundary conditions) of the original recurrence relation fails, you can just remove them from the new equation/inequality and introduce new base case for which the equation/inequality holds.
I don't understand what the paragraph is getting at.
This paragraph is just one simple example demonstrating how a new substitution function fails at the original boundary conditions and how to overcome that problem.
How do we determine the base case in the general case?
There is no a standard way to do that, use you intuition, guess, and check if it works. You should choose your boundary conditions carefully. In the example you in your post $T(2)$ and $T(3)$ are chosen as new boundary conditions since values of $T(2)$ and $T(3)$ directly depend on the value of $T(1)$ which is troublesome. So we just put $T(2)=4$ and $T(3)=5$ without using $T(1)$.
Making a good guess
Unfortunately, there is no general way to guess the correct solutions
to recurrences. Guessing a solution takes experience and,
occasionally, creativity. Fortunately, though, you can use some
heuristics to help you become a good guesser.
Is this acceptable, or would that constitute a sloppy proof?
This inequality should be proven. Assume that our claim is true for all integers less than or equal to $n$. We need to show that the inequality is true for $n+1$.
$$ T(n+1) = 2T((n+1)/2) + (n+1) \leq 2c\frac{n+1}{2}\lg{\frac{n+1}{2}} + (n+1) = c(n+1)\lg{\frac{n+1}{2}} + (n+1)=(n+1)(c\lg{\frac{n+1}{2}} + 1) \leq (n+1)(c\lg{\frac{n+1}{2}} + c) = c(n+1)(\lg{\frac{n+1}{2}} + 1) = c(n+1)(\lg{\frac{n+1}{2}} + \lg2) = c(n+1)\lg{(n+1)}$$
