= Solution
There are two normalization problems in the printed energy identity. In <conformal time>, the physical density of a homogeneous scalar is $T_{00}/a^2$, and its kinetic contribution contains $\phi'^2$, not one power of $\phi'$. The <conformal density of a canonical scalar field> follows directly from the <stress-energy tensor>:
$$
\rho_\phi=\frac{\phi'^2}{2a^2}+V,\qquad
p_\phi=\frac{\phi'^2}{2a^2}-V,\qquad T_{00}=a^2\rho_\phi.
$$
Writing $K=\phi'^2/(2a^2)$, the condition $K-V=-(K+V)/3$ gives $V=2K$, hence
$$
\boxed{K=\rho_\phi/3,\qquad V=2\rho_\phi/3,\qquad
\phi'^2=\frac{2a^2\rho_\phi}{3}=\frac{2T_{00}}3.}
$$
The reconstruction is consistent with this corrected identity; the unsquared printed identity cannot hold as an energy relation.
The <constant-equation-of-state density scaling> for this <coasting fluid> gives $\rho_\phi=\rho_{X0}a^{-2}$, where $\rho_{X0}=3M^2H_0^2\Omega_X$. Choose the branch on which $\phi$ increases. Then
$$
\phi'=\sqrt2MH_0\sqrt{\Omega_X},\qquad
V=\frac{2M^2H_0^2\Omega_X}{a^2}.
$$
Keeping all three density parameters explicitly, the <Friedmann equations> in <conformal time> give
$$
a'^2=H_0^2\left(\Omega_r+\Omega_ma+\Omega_Xa^2\right),\qquad
\frac{da}{d\phi}=\frac1{\sqrt2M}
\sqrt{a^2+\frac{\Omega_m}{\Omega_X}a+\frac{\Omega_r}{\Omega_X}}.
$$
For $\Omega_X>0$, define
$$
b=\frac{\Omega_m}{2\Omega_X},\qquad
A=\frac{\sqrt{\Omega_m^2/4-\Omega_r\Omega_X}}{\Omega_X}.
$$
The square root is $\sqrt{(a+b)^2-A^2}$. Integrating on the expanding branch gives
$$
\operatorname{arcosh}\frac{a+b}{A}
=\frac{\phi-\phi_0}{\sqrt2M}+C,
$$
where $a(\phi_0)=1$. Thus the <scalar-field reconstruction of a coasting component> is
$$
\boxed{a=A\cosh[B(\phi-\phi_0)+C]+D,\quad
B=\frac1{\sqrt2M}=\frac{\sqrt{4\pi}}{m_{\rm pl}},\quad
C=\operatorname{arcosh}\frac{1+b}{A},\quad D=-b.}
$$
Substitution into the potential gives the requested present-field parametrization:
$$
\boxed{V(\phi)=\frac{2M^2H_0^2\Omega_X}
{\left[A\cosh\left(\frac{\phi-\phi_0}{\sqrt2M}+C\right)-b\right]^2}.}
$$
The domain is the branch where $a>0$ and the hyperbolic argument is positive; reversing the direction of the field changes the sign of $B$. The actual discriminant condition is $\Omega_m^2/4>\Omega_r\Omega_X$. The supplied stronger bound suffices for physical density fractions $0<\Omega_X\le1$. No density parameter has been eliminated using the flatness sum.
As a check, $\phi'$ is constant and $V_{,\phi}=-2V(a'/a)/\phi'$. Since $a^2V=\phi'^2$ along this solution, $\phi''+2(a'/a)\phi'+a^2V_{,\phi}=0$ identically. The reconstructed field therefore satisfies its equation of motion as well as the <Friedmann equations> and required <equation of state>.
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