The question rephrased and compressed:
Let $F=F_2[a]$ be a finite extension field of the field of two elements $F_2$. We are given a polynomial $R(X)\in F[X]$, and pairwise coprime irreducible polynomials $m_1(X), m_2(X),\ldots, m_n(X)\in F[X]$. Let $M(X)=\prod_{i=1}^n m_i(X)$ be their product. Is it always possible to find polynomials $P(X)$, $Q_1(X)$ in the ring $F[X]$ such that they satisfy the polynomial congruence $ P(X)\equiv Q_1(X) R(X) \pmod{M(X)} $ as well as the degree constraints $\deg(Q_1) \leq \dfrac{\deg (M)-1}{2}$, $\deg(P) \leq \dfrac{\deg (M)}{2}$?
If the answer is in the affirmative, then how can we efficiently compute the polynomials $b_i(X)$ satisfying the individual congruences $P(X)^2\equiv b_i(X)R(X)^2 \pmod{m_i(X)}, \text{ for }\ i = 1,\ldots,n.$
[The earlier formulation of the question follows.]
Given the following congruences, all polynomials are in $F[X]$ (where $F = \mathbb{F}_2/p(X)$) and $m_i(X)$ coprimes between them $P(X)^2\equiv b_i(X)R(X)^2 \pmod{m_i(X)}, \text{ for } i = 1,\ldots,n.$
Applying CRT, I get $P(X)^2\equiv{Q(X)R(X)^2}\pmod{M(X)},$
where $Q(X) = \sum_{i=1}^n{b_i(X)y_i(X)M_i(X)}\, $ with $M_i(X) = M(X)/m_i(X)$ and $y_i(X)$ is a multiplicative inverse of $M_i(X)$ module $m_i(X)$; $M(X) = \prod_{i=1}^{n}{m_i(X)}$
and resolving this quadratic equation I get:
$P(X)\equiv Q_1(X)R(X) \pmod{M(X)}$
For example:
For $K = \mathbb{F}_2, F = K/(X^4+X+1)$ and $a$ primitive element of $F$
Give:
$M = (X + 1) \cdot (X + a^2 + a + 1) \cdot (X + a^3 + a + 1) \cdot (X + a^3 + a^2)$
$\eqalign{R &= a^2 X^3 + (a^3 + a) X^2 + (a^3 + 1) X\cr}$
Applying CRT (with $b_i$ than variables) get
$\eqalign{QR^2&=(a^3X^5 + a^3X^4 + a^3X^3 + a^{14}X^2 + a^{13}X + a^5) \cdot R^2 \cdot b_1 \cr & + (a^{12}X^5 + a^7X^4 + a^{12}X^3 + a^9X^2 + a^2X + a^8) \cdot R^2 \cdot b_2 \cr & + (a^5X^5 + a^{12}X^4 + a^5X^3 + a^{11}X^2 + a^9X + a^7) \cdot R^2 \cdot b_3 \cr & + (X^5 + a^6X^4 + X^3 + a^{13} X^2 + a X + a^4) \cdot R^2 \cdot b_4}$,
and obviusly I don't resolved this quadratic equation $P^2\equiv QR^2 \pmod{M},$, because the $b_i$'s are unknow values, but I will expect this expression :
$P\equiv{Q_1R}\pmod{M}$
My question: Let $M(X)$ and $R(X)$ is there any mode to choose $b_i$ and $P(X)$ where $\deg(Q_1) < \dfrac{\deg (M)-1}{2}$ and $R Q_1$ have this form $MK+P$ (obviously) and $\deg(P) < \dfrac{\deg (M)}{2}$ ?