The $cs$ terms is present in real partial fraction decomposition. You are decomposing into proper rational functions, but when the denominator has degree $2$, your numerator must have degree $1$ to represent all possible proper rational functions. The correct form for a general factor $a x^n + \cdots + b$ of the denominator is
$$ \frac{A x^{n-1} + B x^{n-2} + \cdots + C x + D}{a x^n + \cdots + b} \text{,} $$
that is, the numerator is always one degree less then the factor in the denominator. (If the factor appears to a power, the numerator still gets its degree from the factor; it ignores the power.)
This is demonstrated on the partial fractions page as example 2.
More pointedly, say one needs to decompose
$$ \frac{x^2}{x(x^2-4x+8)} $$
There is no solution in reals $A$ and $B$ such that
$$ \frac{x^2}{x(x^2-4x+8)} = \frac{A}{x} + \frac{B}{x^2-4x+8} \text{.} $$
When you clear the denominators, to match the quadratic coefficients, $A = 1$, but there is no choice of $B$ such that $-4x+8+Bx = 0$. For a solution to exist, you need
$$ \frac{x^2}{x(x^2-4x+8)} = \frac{A}{x} + \frac{Bx + C}{x^2-4x+8} $$
to arrive at $-4x+8+Bx + C = 0$, which does have a solution.
Now, as to your particular problem ... Clearing denominators, we have \begin{align*}
2s^2+5s+12 &= \frac{a (s^2+2s+10)(s+2)}{s+2}+\frac {(b+cs)(s^2+2s+10)(s+2)}{s^2+2s+10} \\
&= a (s^2+2s+10) + (b+cs)(s+2) \\
&= (as^2 + 2as + 10a) + (b(s+2)+cs(s+2)) \\
&= (as^2 + 2as + 10a) + (bs+2b+cs^2 + 2cs) \\
&= (a+c) s^2 + (2a + b + 2c)s + (10a + 2b) \text{.}
\end{align*}
For two polynomials (real or complex) to agree, they must agree in each coefficient, so we have the system
\begin{align*}
a + c &= 2 \\
2a+b+2c &= 5 \\
10a+2b &= 12
\end{align*}
Subtracting two copies of the second from the first, $b = 1$. Using that in the third, $a = 1$. Using that in the first, $c = 1$. Therefore,
$$ \frac {2s^2+5s+12}{(s+2)(s^2+2s+10)} = \frac{1}{s+2} + \frac{s+1}{s^2+2s+10} \text{.} $$
That's the real decomposition. Continuing, using the complex factoring
$$ s^2+2s+10 = (s - (-1-3\mathrm{i}))(s - (-1+3\mathrm{i})) \text{,} $$
$$ \frac{s+1}{s^2+2s+10} = \frac{d}{s - (-1-3\mathrm{i}))} + \frac{e}{s - (-1+3\mathrm{i}))} \text{.} $$
Clearing denominators, we obtain
\begin{align*}
s+1 &= \frac{d(s^2+2s+10)}{s - (-1-3\mathrm{i}))} + \frac{e(s^2+2s+10)}{s - (-1+3\mathrm{i}))} \\
&= d(s - (-1+3\mathrm{i})) + e(s - (-1-3\mathrm{i})) \\
&= ds - d(-1+3\mathrm{i})) + es - e(-1-3\mathrm{i})) \\
&= (d+e)s +d -3d\mathrm{i} + es + e +3e\mathrm{i} \\
\end{align*}
Again equating coefficients, \begin{align*}
d+e &= 1 \\
d+e + 3(-d+e)\mathrm{i} &= 1 + 0 \mathrm{i} \text{.}
\end{align*}
Subtracting the first of these from the second, $3(-d+e)\mathrm{i} = 0$, forcing $d = e$. Then the first gives $d = e = 1/2$.
The development above first expanded in factors of the denominator that were reducible over the reals, then factored further over the complexes. These can be done all at once.
$$ \frac{2s^2+5s+12}{(s^2+2s+10)(s+2)} = \frac{a }{s+2}+\frac{d}{s - (-1-3\mathrm{i}))} + \frac{e}{s - (-1+3\mathrm{i}))} \text{.} $$
Now all the denominators are (powers of) factors of degree $1$, so their numerators have degree $0$.
