The problem as stated is unsolvable in the case of $\mathbb{Z}$ coefficients with $f(a)$ a prime power. Proof: Let $a=0$, let $f(0) = p^n$ and let the coefficient of $x$ in $f$ not be divisible by $p$. Then $p(0)$, $q(0)$, $r(0)$ and $r(0)$ are all of the form $\pm p^k$, and the hypotheses forbid that $k=0$. So all of the constant terms of $p(x)$, $q(x)$, $r(x)$ and $s(x)$ are divisible by $p$. But then the linear term of $f(x) = (p(x)q(x)+r(x)s(x))/2$ is divisible by $p$ as well, a contradiction. $\square$
Here is a solution for $\mathbb{Z}$ coefficients, assuming that $f(a)$ is not a prime power.
Adapting it for coefficients in a finite field should be easy, since there are fewer things to worry about.
Without loss of generality, let $a=0$. Since $f(0)$ is not a prime power, we can factor it as factor $f(0)$ as $p_0 \cdot r_0$ with $GCD(p_0,r_0)=1$. Find $c$ and $d$ such that $c p_0 -d r_0=1$, note for future reference that this implies $GCD(c,d)=1$. We will take $p(x) = dx+p_0$ and $r(x) = cx+r_0$, and slowly construct $q(x)$ and $s(x)$.
Step 1 Making $2 f(x) = p(x) q_0(x) + r(x) s_0(x)$.
We just need to show that $2 f(x)$ is in the ideal generated by $p(x)$ and $r(x)$. But $c p(x) - d r(x) = c (dx+p_0) - d (cx+r_0) = c p_0 - d r_0 = 1$, so this ideal is the whole ring and every polynomial is in it. Explicitly, $2 f(x) = (2 c f(x)) p(x) - (2 d f(x)) r(x)$. $\square$
Step 2 Making $2 f(x) = p(x) q_1(x) + r(x) s_1(x)$ with $\deg q_1$ and $\deg s_1 \leq \deg f -1$.
Suppose that we have found polynomial $Q(x)$ and $S(x)$ with $2 f(x) = p(x) Q(x) + r(x) S(x)$ and $\deg Q$ or $\deg R \geq \deg f$. We will show that we can replace them by polynomials $Q'(x)$ and $S'(x)$ of lower degree, still obeying the equation $2 f(x) = p(x) Q'(x) + r(x) S'(x)$.
Since we are supposing that at least one of $p(x) Q(x)$ and $r(x) S(x)$ has degree larger than the sum $p(x) Q(x) + r(x) S(x)$, we see that the leading terms must cancel. Let the leading terms of $Q$ and $S$ be $Q_m x^m + \cdots$ and $S_m x^m + \cdots$. So $d Q_m + c S_m=0$. Using our above observation that $GCD(c,d)=1$, we see that $Q_m = c F$ and $S_m = -d F$ for some integer $F$. Replace $Q(x)$ and $S(x)$ by $Q(x) - F r(x) x^{m-1}$ and $S(x) + F p(x) x^{m-1}$. $\square$.
Step 3 Achieving $2 f(x) = p(x) q(x) + r(x) s(x)$ with $p(0) q(0) = r(0) s(0)$, and $\deg q$, $\deg s \leq \deg f-1$.
At this point in the proof we already have $2 f(x) = p(x) q_1(x) + r(x) s_1(x)$ with $q_1$ and $s_1$ of sufficiently low degree; we just must change their values at $0$ without making their degree larger.
Comparing constant terms,
$$2 f(0) = 2 p_0 r_0 = p_0 q_1(0) + r_0 s_1(0).$$
Using $GCD(p_0, r_0) = 1$, we see that $r_0 | q_1(0)$ and $p_0 | s_1(0)$. Let $q_1(0) = r_0 (1-G)$ and $s_1(0) = p_0 (1+G)$ for some integer $G$. Then replace $q_1(x)$ and $s_1(x)$ by $q_1(x) + G r(x)$ and $s_1(x) - G p(x)$. Since $\deg p = \deg r =1 \leq \deg f-1$, this replacement can't make the degrees of $q$ and $s$ larger than $\deg f-1$. $\square$.
Where did this problem come from? It would make a nice Putnam problem (not now, of course).
