The key here is to look at the condition $g(x) = cx^2 + h(x)\qquad\mbox{with $\lim\limits_{x\to 0}\frac{h(x)}{x^2} = 0$.}$
If you think about a potential Taylor expansion of $g(x)$ of degree $2$, this tells you that $h(x)$ has to consists of the degree $3$ and above terms only; that means that the Taylor expansion of $g(x)$ around $0$ cannot have a linear or constant terms, so $g(0)=0$ and g'(0)=0.
Now, since we will want $g(x) = f(x+a)+b$, then g'(x) = f'(x+a), so in order to get g'(0)=0, we need f'(a)=0. The only point where we know f'(a)=0 is at $a=x_0$, so we'll pick $a=x_0$. Thus, $g(x) = f(x+x_0) + b$. In order to also get $g(0) = 0$, we need $f(x_0) + b = 0$, so $b=-f(x_0)$. Thus, the function we expect will work is $g(x) = f(x+x_0) - f(x_0).$ Now, to see this works, we proceed as you did: taking the Taylor expansion for $f$ around $x_0$ up to degree $2$, we get f(x) = f(x_0) + \frac{f''(x_0)}{2}(x-x_0)^2 + R_3(x). So g(x) = f(x+x_0) - f(x_0) = \frac{f''(x_0)}{2}x^2 + R_3(x+x_0). Thus, c=\frac{f''(x_0)}{2}, and $h(x) = R_3(x+x_0)$. We just need to show that $h(x)/x^2 \to 0$ as $x\to 0$, but this is equivalent to $R_3(x+x_0)/x^2\to 0$ as $x\to 0$, which in turn is equivalent to $\lim_{x\to x_0}\frac{R_3(x)}{(x-x_0)^2} = 0.$ This hold precisely because $R_3$ is the degree $3$ remainder at $x_0$.
So, in summary, we set $a=x_0$, $b=-f(x_0)$, c=\frac{f''(x_0)}{2}, $g(x) = f(x+x_0) - f(x_0)$, and $h(x) = R_3(f,x_0,x)$ (the Remainder of degree $3$ in the Taylor expansion of $f$ around $x_0$, evaluated at $x$).