Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder is linear?

Because by definition the quotient and the remainder of the division of a polynomial $p_1(x)$ by a polynomial $p_2(x)$ are polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that

1. $p_1(x)=p_2(x)q(x)+r(x)$;
2. $r(x)=0$ or its degree is smaller than the degree of $p_2(x)$.

In particular, if $p_2(x)$ is a quadratic polynomial, then the degree of $r(x)$ will be at most $1$.

While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)

If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.

For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:

1. Note that the ratio of the leading terms is $\frac {3x^3}{x^2} = 3x$
2. Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$
3. Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $\frac {-2x^2}{x^2} = -2$.
4. Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$
5. Note that the ratio of the leading terms is now $\frac {-2x}{x^2} = \frac {-2}x$, which is not a polynomial, so we cannot continue.

Combining the results from steps 2 and 4: $$\begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\\&=(3x - 2)(x^2 + 1) + (-2x-3)\end{align}$$

If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.

The point is we can prove as much. Let the quadratic be $q:=ax^2+bx+c$. Working modulo $q$, $x^2=-(bx+c)/a$. Each $x^n$ is then of the form $cx+d$ modulo $q$; you can prove this by induction (you can even get recursion relations on the coefficients). For example, $x^3=-(bx^2+cx)/a$, but we can then replace $x^2$ as above.