I'm posting an "answer" for alternative methods. The third method below is the most straight forward, exploiting the fact that $(x^2+6)-(x^2+4) = 2$. The answer to the question, r(4) = -131.
Using a "reverse" long division process to produce a common f(x) based on the first two given equations will work, but, although this solves the problem, I doubt that is the intended solution, since it involves a reasoned trial and error search for f(x) (sort of an optimized brute force search), and it is my impression that a proper answer should be able to solve for r(x) or specifically for r(4) without having to determine f(x).
Below is what the process looks like. f(x) (the dividend) and the quotients a(x), b(x) are unknown. The divisors and remainders are given in the first two equations of the question. You start at the bottom of these two long hand divisions in parallel, working upwards to produce a common f(x).
As mentioned, this is a reasoned trial an error process. For example, my first attempt at the x^2 term of f(x) was 13x^2, which failed later, the second attempt was 25x^2, which worked (at least it produces a common f(x) that satisfies the first two equations). For the rest of the terms, the first attempts at terms for a common f(x) (and the corresponding quotient terms of a(x) and b(x)) worked.
Consider the very first step, f(x) / (x^2+4) has remainder ...+1, f(x) / (x^2+6) has remainder ...-1. This suggests that the last term of f(x) is 5 and the last terms of both quotients are 1, since 5-4 = +1 and 5-6 = -1. The x terms in the remainder show that after subtraction from the third step from the bottom, x terms are 2x for division by (x^2+4) and 6x for division by (x^2+6), and setting the x term of f(x) to 18 works as 18 - (4·4) = 2 and 18 - (2·6) = 6. The process is continued upwards, looking for common f(x) terms that satisfy both long hand divisions. This is the final result. Again note that this process is started at the bottom and worked upwards to produce a common f(x) (dividend) for both divisors:
1 1 6 4 1 1 1 4 2 1
--------------------- ---------------------
1 0 4 | 1 1 10 8 25 18 5 1 0 6 | 1 1 10 8 25 18 5
1 0 4 1 0 6
1 6 8 1 4 8
1 0 4 1 0 6
6 4 25 4 2 25
6 0 24 4 0 24
4 1 18 2 1 18
4 0 16 2 0 12
1 2 5 1 6 5
1 0 4 1 0 6
2 1 6 -1
Once any f(x) is determined that satisfies the first two given equations, then the rest just requires normal division.
$$f(x) = x^6 + x^5 + 10 x^4 + 8 x^3 + 25 x^2 + 18 x + 5$$
Expressing f(x) as quotient · divisor + remainder for the different divisors:
$$f(x) = (x^4 + x^3 + 6x^2 + 4x + 1)(x^2 + 4) + 2x + 1 $$
$$f(x)= (x^4 + x^3 + 4x^2 + 2x + 1)(x^2 + 6) + 6x - 1 $$
$$f(x) = (x^2 + x)(x^4 + 10x^2 + 24) -2 x^3 + x^2 - 6 x + 5 $$
Using typical remainder theorem approach
f(x) evaluated at the 4 roots of (x^2+4)(x^2+6):
$$f(x) = c(x))(x^2+4)(x^2+6)) = r(x)$$
$$f(x) = (c(x) · 0) + r(x) = r(x)$$
f(x) evaluated at the 2 roots of (x^2+4):
$$f(x) = (a(x) · 0) + 2x + 1) = 2x + 1$$
f(x) evaluated at the 2 roots of (x^2+6):
$$f(x) = (b(x) · 0) + 6x - 1) = 6x - 1$$
This leads to 4 data points for r(x):
$${-(2)i,-(4)i+1}$$
$${+(2)i,+(4)i+1}$$
$${-\sqrt{6}i,-(6)\sqrt{6}i-1}$$
$${-\sqrt{6}i,-(6)\sqrt{6}i-1}$$
Using Lagrange interpolation to solve for r(x) is complicated due to the complex numbers:
r(x) = ((x-x1)(x-x2)(x-x3)(y0))/((x0-x1)(x0-x2)(x0-x3)) +
((x-x0)(x-x2)(x-x3)(y1))/((x1-x0)(x1-x2)(x1-x3)) +
((x-x0)(x-x1)(x-x3)(y2))/((x2-x0)(x2-x1)(x2-x3)) +
((x-x0)(x-x1)(x-x2)(y3))/((x3-x0)(x3-x1)(x3-x2)) +
grinding the 4 terms leads to:
r(x) = (1/2 + i/8) (x^3 - 2 i x^2 + 6 x - 12 i) +
(1/2 - i/8) (x^3 + 2 i x^2 + 6 x + 12 i) +
1/24 (( i sqrt(6) - 36) x + 36 i sqrt(6) + 6) (x^2 + 4) +
1/24 ((-i sqrt(6) - 36) x - 36 i sqrt(6) + 6) (x^2 + 4)
r(x) = x^3 + 1/2 x^2 + 6 x + 3 +
-3 x^3 + 1/2 x^2 - 12 x + 2
r(x) = -2 x^3 + x^2 - 6 x + 5
Exploiting the fact that $(x^2+6) - (x^2+4) = 2$ :
$$f(x) = a(x)(x^2+4)+(2x+1)$$
$$f(x) = b(x)(x^2+6)+(6x-1)$$
multiply 1st equation by $(x^2+6)$ and 2nd equation by $(x^2+4)$
$$f(x)(x^2+6) = a(x)(x^2+4)(x^2+6)+(2x+1)(x^2+6)$$
$$f(x)(x^2+4) = b(x)(x^2+4)(x^2+6)+(6x-1)(x^2+4)$$
subtracting 4th equation from 3rd:
$$2f(x) = a(x)(x^2+4)(x^2+6)+(2x+1)(x^2+6)-b(x)(x^2+4)(x^2+6)-(6x-1)(x^2+4)$$
$$f(x) = (1/2)(a(x)(x^2+4)(x^2+6)+(2x+1)(x^2+6)-b(x)(x^2+4)(x^2+6)-(6x-1)(x^2+4))$$
$$f(x) mod((x^2+4)(x^2+6)) = r(x) = (1/2)((2x+1)(x^2+6) - (6x-1)(x^2+4))$$
$$r(x) = -2x^3 + x^2 - 6x + 5$$
