Simply explaining: the key IDEA is that you are not done dividing until your remainder is not further divisible by your divisor. The key WORD in your post is "GENERALLY" because the conclusion is not hard and fast, it is simply "usually" so.
IDEA first. Consider simple numbers, perhaps dividing $1\,034$ by $10$. You begin at the left and work right (this is taught in first grade of course so I won't belabor the process) until you cannot usefully continue. So you get $100~R~34$, and $3~R~4$, and stop when left with a remainder ($4$) less than your divisor, $10$. Easy enough to figure when it's "less than" with numbers, but slightly less so with polynomials. But we see it slightly differently with polynomials: we stop when the order of the remainder is lower than the order of the divisor. So dividing by $7x^2$, and having a remainder with just $x^2$, we usually grit our teeth (we DO love our integers…) and use $1/7$ eliminating the remainder's $2$nd order term ($x^2$). At that point, coninuing becomes difficult even with gritted teeth.
And so we stop. Notice that it is not that the remainder is now linear with that being special in some way: it is simply of lower order than the divisor. In the case of dividing by a $2$nd order polynomial, that means it is $1$st order, or simply a constant (number).
Which brings in the "generally" part: in practice, it COULD be just the constant, and not a linear equation. $x^2 + 8$ divided by $x^2$, for example. But not usually. Hence "generally" in your post's statement. And moving up in order, say a fifth order divisor, while you'd usually expect, just out of probability, a fourth order remainder, it could also be a third, second, or first order remainder, or just a constant.
So it's not a great theorem with a stunning proof, just a practical observation based upon the mechanics we use for the division.
(And it adds an "idiot check" in case one stops early, thinking he's finished. "Wait, I can't be, it ought to be linear and it isn't so maybe I need to keep working…")