Because your formulae for the pion wave functions are written in a confusing way. The quark composition of the positively charged pion is
$$ |\pi^+\rangle = |u\bar d\rangle $$
while the neutral pion is
$$ |\pi^0\rangle = \frac{|u\bar u\rangle - |d\bar d\rangle}{\sqrt{2}}. $$
These two composite formulae are completely analogous: these pions are mesons, i.e. quark-antiquark bound states, and they just differ in which quarks from the double $(u,d)$ are used. The neutral pion is a superposition of two pieces: superpositions of these types are omnipresent everywhere in quantum theory. The charged pions only contain one term because there is only one quark-antiquark combination involving $u,d$ quarks and antiquarks whose total charge is $\pm 1$. To get $Q=0$, one may either combine $u\bar u$ or $d\bar d$ and the relative complex coefficient between these two terms is a priori arbitrary and determines whether the superposition is a mass eigenstate and if it is, whether it's lighter or heavier.
Of course that the detailed formulae for the two pions must be different in some respects, otherwise they would be an identical particle.
During the decay, all the conservation laws and other laws of physics are satisfied.
Note that in the 3-dimensional complex space spanned by basis $\boldsymbol{\lbrace}\boldsymbol{u}\overline{\boldsymbol{u}},\boldsymbol{d}\overline{\boldsymbol{d}},\boldsymbol{s}\overline{\boldsymbol{s}}\boldsymbol{\rbrace}$, this basis is replaced by $\boldsymbol{\lbrace}\boldsymbol{\pi^{0},\boldsymbol{\eta},\boldsymbol{\eta}^{\prime}}\boldsymbol{\rbrace}$ through a special unitary transformation $\mathrm{V}\in SU(3)$,
\begin{equation}
\begin{bmatrix}
\boldsymbol{\pi^{0}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{\eta} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{\eta}^{\prime} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\sqrt{\tfrac{1}{2}} & \boldsymbol{-} \sqrt{\tfrac{1}{2}} & \hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\
\sqrt{\tfrac{1}{6}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{6}} & \boldsymbol{-}\sqrt{\tfrac{2}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}\\
\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} & \hphantom{\boldsymbol{-}}\sqrt{\tfrac{1}{3}} \vphantom{\tfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
=\mathrm{V}
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{d}\overline{\boldsymbol{d}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}\\
\boldsymbol{s}\overline{\boldsymbol{s}} \vphantom{\dfrac{a}{\tfrac{a}{b}}}
\end{bmatrix}
\tag{1}\label{1}
\end{equation}
see Figure.
$
\newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}}
\newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}}
\newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}}
\newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}}
\newcommand{\MM}[4]
{\begin{bmatrix}
#1 & #2\\
#3 & #4\\
\end{bmatrix}}
\newcommand{\MMM}[9]
{\textstyle \begin{bmatrix}
#1 & #2 & #3 \\
#4 & #5 & #6 \\
#7 & #8 & #9 \\
\end{bmatrix}}
\newcommand{\CMRR}[2]
{\begin{bmatrix}
#1 \\
#2
\end{bmatrix}}
\newcommand{\CMRRR}[3]
{\begin{bmatrix}
#2 \\
#3
\end{bmatrix}}
\newcommand{\CMRRRR}[4]
{\begin{bmatrix}
#1 \\
#2 \\
#3 \\
#4
\end{bmatrix}}
\newcommand{\RMCC}[2]
{\begin{bmatrix}
#1 & #2
\end{bmatrix}}
\newcommand{\RMCCC}[3]
{\begin{bmatrix}
#1 & #2 & #3
\end{bmatrix}}
\newcommand{\RMCCCC}[4]
{\begin{bmatrix}
#1 & #2 & #3 & #4
\end{bmatrix}}
$
$\boldsymbol{\S\:}\textbf{A. Mesons from three quarks}$ $\boldsymbol{u},\boldsymbol{d},\boldsymbol{s} : \boldsymbol{3}\boldsymbol{\otimes}\overline{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{8}$
Suppose we know the existence of three quarks only : $\boldsymbol{u}$, $\boldsymbol{d}$ and $\boldsymbol{s}$. Under full symmetry these are the basic states, let
\begin{equation}
\boldsymbol{u}=
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
\qquad
\boldsymbol{d}=
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\qquad
\boldsymbol{s}=
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\tag{001}\label{001}
\end{equation}
of a 3-dimensional complex Hilbert space of quarks, say $\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$. A quark $\boldsymbol{\xi} \in \mathbf{Q}$ is expressed in terms of these basic states as
\begin{equation}
\boldsymbol{\xi}=\xi_u\boldsymbol{u}+\xi_d\boldsymbol{d}+\xi_s\boldsymbol{s}=
\begin{bmatrix}
\xi_u\\
\xi_d\\
\xi_s
\end{bmatrix}
\qquad \xi_u,\xi_d,\xi_s \in \mathbb{C}
\tag{002}\label{002}
\end{equation}
For a quark $\boldsymbol{\zeta} \in \mathbf{Q}$
\begin{equation}
\boldsymbol{\zeta}=\zeta_u\boldsymbol{u}+\zeta_d\boldsymbol{d}+\zeta_s\boldsymbol{s}=
\begin{bmatrix}
\zeta_u\\
\zeta_d\\
\zeta_s
\end{bmatrix}
\tag{003}\label{003}
\end{equation}
the respective antiquark $\overline{\boldsymbol{\zeta}}$ is expressed by the complex conjugates of the coordinates
\begin{equation}
\overline{\boldsymbol{\zeta}}=\overline{\zeta}_u \overline{\boldsymbol{u}}+\overline{\zeta}_d\overline{\boldsymbol{d}}+\overline{\zeta}_s\overline{\boldsymbol{s}}=
\begin{bmatrix}
\overline{\zeta}_u\\
\overline{\zeta}_d\\
\overline{\zeta}_s
\end{bmatrix}
\tag{004}\label{004}
\end{equation}
with respect to the basic states
\begin{equation}
\overline{\boldsymbol{u}}=
\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}
\qquad
\overline{\boldsymbol{d}}=
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
\qquad
\overline{\boldsymbol{s}}=
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\tag{005}\label{005}
\end{equation}
the antiquarks of $\boldsymbol{u},\boldsymbol{d}$ and $\boldsymbol{s}$ respectively. The antiquarks belong to a different space, the space of antiquarks $\overline{\mathbf{Q}}\equiv \mathbb{C}^{\boldsymbol{3}}$.
Since a meson is a quark-antiquark pair, we'll try to find the product space
\begin{equation}
\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\: \left(\equiv \mathbb{C}^{\boldsymbol{9}}\right)
\tag{006}\label{006}
\end{equation}
Using the expressions \eqref{002} and \eqref{004} of the quark $\boldsymbol{\xi} \in \mathbf{Q}$ and the antiquark $\overline{\boldsymbol{\zeta}} \in \overline{\mathbf{Q}}$ respectively, we have for the product meson state $ \mathrm{X} \in \mathbf{M}$
\begin{equation}
\begin{split}
\mathrm{X}=\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_u\overline{\zeta}_d \left( \boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_u\overline{\zeta}_s \left(\boldsymbol{u}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+ \\
&\xi_d\overline{\zeta}_u \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_d\overline{\zeta}_d \left( \boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_d\overline{\zeta}_s \left(\boldsymbol{d}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)+\\
&\xi_s\overline{\zeta}_u \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{u}}\right)+\xi_s\overline{\zeta}_d \left( \boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{d}}\right)+\xi_s\overline{\zeta}_s \left(\boldsymbol{s}\boldsymbol{\otimes}\overline{\boldsymbol{s}}\right)
\end{split}
\tag{007}\label{007}
\end{equation}
In order to simplify the expressions, the product symbol $"\boldsymbol{\otimes}"$ is omitted and so
\begin{equation}
\begin{split}
\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=&\xi_u\overline{\zeta}_u \boldsymbol{u}\overline{\boldsymbol{u}}+\xi_u\overline{\zeta}_d \boldsymbol{u}\overline{\boldsymbol{d}}+\xi_u\overline{\zeta}_s \boldsymbol{u}\overline{\boldsymbol{s}}+ \\
&\xi_d\overline{\zeta}_u \boldsymbol{d}\overline{\boldsymbol{u}}+\xi_d\overline{\zeta}_d \boldsymbol{d}\overline{\boldsymbol{d}}+\xi_d\overline{\zeta}_s \boldsymbol{d}\overline{\boldsymbol{s}}+\\
&\xi_s\overline{\zeta}_u \boldsymbol{s}\overline{\boldsymbol{u}}+\xi_s\overline{\zeta}_d \boldsymbol{s}\overline{\boldsymbol{d}}+\xi_s\overline{\zeta}_s \boldsymbol{s}\overline{\boldsymbol{s}}
\end{split}
\tag{008}\label{008}
\end{equation}
Due to the fact that $\mathbf{Q}$ and $\overline{\mathbf{Q}}$ are of the same dimension, it's convenient to represent the meson states in the product 9-dimensional complex space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ by square $3 \times 3$ matrices instead of row or column vectors
\begin{equation}
\mathrm{X}=\boldsymbol{\xi}\overline{\boldsymbol{\zeta}}=
\begin{bmatrix}
\xi_u\overline{\zeta}_u & \xi_u\overline{\zeta}_d & \xi_u\overline{\zeta}_s\\
\xi_d\overline{\zeta}_u & \xi_d\overline{\zeta}_d & \xi_d\overline{\zeta}_s\\
\xi_s\overline{\zeta}_u & \xi_s\overline{\zeta}_d & \xi_s\overline{\zeta}_s
\end{bmatrix}=
\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u \\
\overline{\zeta}_d \\
\overline{\zeta}_s
\end{bmatrix}^{\mathsf{T}}
=
\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s
\end{bmatrix}
\tag{009}\label{009}
\end{equation}
The product space $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is created by completion of the set of states \eqref{008} with arbitrary complex coefficients
\begin{equation}
\begin{split}
\mathrm{X}=&\mathrm{x}_{_{11}}\boldsymbol{u}\overline{\boldsymbol{u}}+\mathrm{x}_{_{12}} \boldsymbol{u}\overline{\boldsymbol{d}}+\mathrm{x}_{_{13}} \boldsymbol{u}\overline{\boldsymbol{s}}+ \\
&\mathrm{x}_{_{21}}\boldsymbol{d}\overline{\boldsymbol{u}}+\mathrm{x}_{_{22}} \boldsymbol{d}\overline{\boldsymbol{d}}+\mathrm{x}_{_{23}} \boldsymbol{d}\overline{\boldsymbol{s}}+ \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}\\
&\mathrm{x}_{_{31}} \boldsymbol{s}\overline{\boldsymbol{u}}+\mathrm{x}_{_{32}} \boldsymbol{s}\overline{\boldsymbol{d}}+\mathrm{x}_{_{33}} \boldsymbol{s}\overline{\boldsymbol{s}}
\end{split}
\tag{010}\label{010}
\end{equation}
that is
\begin{equation}
\mathrm{X}=
\begin{bmatrix}
\mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\
\mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\
\mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}}
\end{bmatrix} \:, \qquad \mathrm{x}_{_{ij}} \in \mathbb{C}
\tag{011}\label{011}
\end{equation}
So $\:\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\:$ is identical to $\mathbb{C}^{\boldsymbol{9}}$ with base states
\begin{align}
&\boldsymbol{u}\overline{\boldsymbol{u}}=
\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{u}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{u}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\tag{012a}\label{012a}\\
&\boldsymbol{d}\overline{\boldsymbol{u}}=
\begin{bmatrix}
0 & 0 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\boldsymbol{d}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}
\quad
\:\boldsymbol{d}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}
\tag{012b}\label{012b}\\
&\boldsymbol{s}\overline{\boldsymbol{u}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{bmatrix}
\quad
\:\boldsymbol{s}\overline{\boldsymbol{d}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 1 & 0
\end{bmatrix}
\quad
\:\boldsymbol{s}\overline{\boldsymbol{s}}=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}
\tag{012c}\label{012c}
\end{align}
This basis is represented symbolically by a $3\times 3$ array
\begin{equation}
\mathcal{F}_{\mathbf{M}}=
\begin{bmatrix}
\boldsymbol{u}\overline{\boldsymbol{u}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\
\boldsymbol{d}\overline{\boldsymbol{u}} & \boldsymbol{d}\overline{\boldsymbol{d}} & \boldsymbol{d}\overline{\boldsymbol{s}}\\
\boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & \boldsymbol{s}\overline{\boldsymbol{s}}
\end{bmatrix}
\tag{013}\label{013}
\end{equation}
In this Hilbert space the usual inner product between states
\begin{equation}
\mathrm{X}=
\begin{bmatrix}
\mathrm{x}_{_{11}} & \mathrm{x}_{_{12}} & \mathrm{x}_{_{13}}\\
\mathrm{x}_{_{21}} & \mathrm{x}_{_{22}} & \mathrm{x}_{_{23}}\\
\mathrm{x}_{_{31}} & \mathrm{x}_{_{32}} & \mathrm{x}_{_{33}}
\end{bmatrix} \:, \qquad
\mathrm{Y}=
\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}
\tag{014}\label{014}
\end{equation}
is
\begin{equation}
\begin{split}
\langle \mathrm{X},\mathrm{Y}\rangle \equiv &\mathrm{x}_{_{11}}\overline{\mathrm{y}}_{_{11}}+\mathrm{x}_{_{12}}\overline{\mathrm{y}}_{_{12}}+\mathrm{x}_{_{13}}\overline{\mathrm{y}}_{_{13}}+\\
&\mathrm{x}_{_{21}}\overline{\mathrm{y}}_{_{21}}+\mathrm{x}_{_{22}}\overline{\mathrm{y}}_{_{22}}+\mathrm{x}_{_{23}}\overline{\mathrm{y}}_{_{23}}+\\
&\mathrm{x}_{_{31}}\overline{\mathrm{y}}_{_{31}}+\mathrm{x}_{_{32}}\overline{\mathrm{y}}_{_{32}}+\mathrm{x}_{_{33}}\overline{\mathrm{y}}_{_{33}}
\end{split}
\tag{015}\label{015}
\end{equation}
which, using the $3\times 3$ matrix representation of states, is the trace of the matrix product $\mathrm{X}\BoldExp{\mathrm{Y}}{*}$
\begin{equation}
\langle \mathrm{X},\mathrm{Y}\rangle =\mathrm{Tr}\left[\mathrm{X}\BoldExp{\mathrm{Y}}{*}\right]
\tag{016}\label{016}
\end{equation}
given that $\BoldExp{\mathrm{Y}}{*}$ is the complex conjugate of the transpose of $\mathrm{Y}$
\begin{equation}
\BoldExp{\mathrm{Y}}{*}\equiv
\BoldExp{
\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}}
{*}
=
\overline{\begin{bmatrix}
\mathrm{y}_{_{11}} & \mathrm{y}_{_{12}} & \mathrm{y}_{_{13}}\\
\mathrm{y}_{_{21}} & \mathrm{y}_{_{22}} & \mathrm{y}_{_{23}}\\
\mathrm{y}_{_{31}} & \mathrm{y}_{_{32}} & \mathrm{y}_{_{33}}
\end{bmatrix}^{\mathsf{T}}}
=
\begin{bmatrix}
\overline{\mathrm{y}}_{_{11}} & \overline{\mathrm{y}}_{_{21}} & \overline{\mathrm{y}}_{_{31}}\\
\overline{\mathrm{y}}_{_{12}} & \overline{\mathrm{y}}_{_{22}} & \overline{\mathrm{y}}_{_{32}}\\
\overline{\mathrm{y}}_{_{13}} & \overline{\mathrm{y}}_{_{23}} & \overline{\mathrm{y}}_{_{33}}
\end{bmatrix}
\tag{017}\label{017}
\end{equation}
Now, under a unitary transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\;\mathbf{Q}\;$, we have
\begin{equation}
\BoldExp{\boldsymbol{\xi}}{'} = W\boldsymbol{\xi}
\tag{018}\label{018}
\end{equation}
so in the space of antiquarks $\overline{\mathbf{Q}}\;$, since $\;\BoldExp{\boldsymbol{\zeta}}{'}=W \boldsymbol{\zeta}\;$
\begin{equation}
\overline{\BoldExp{\boldsymbol{\zeta}}{'}}= \overline{W}\;\overline{\boldsymbol{\zeta}}
\tag{019}\label{019}
\end{equation}
and for the meson state
\begin{align}
\BoldExp{\mathrm{X}}{'} & =\BoldExp{\boldsymbol{\xi}}{'}\boldsymbol{\otimes}\overline{\BoldExp{\boldsymbol{\zeta}}{'}}=\left(W\boldsymbol{\xi}\vphantom{\overline{W}\overline{\boldsymbol{\zeta}} }\right)\left(\overline{W}\overline{\boldsymbol{\zeta}} \right)
=
\Biggl(W\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}\Biggr)
\Biggl(\overline{W}\begin{bmatrix}
\overline{\zeta}_u\\
\overline{\zeta}_d\\
\overline{\zeta}_s
\end{bmatrix}\Biggr)^{\mathsf{T}}
\nonumber\\
& = W\Biggl(\begin{bmatrix}
\xi_u\vphantom{\overline{\zeta}_u}\\
\xi_d\vphantom{\overline{\zeta}_u}\\
\xi_s\vphantom{\overline{\zeta}_u}
\end{bmatrix}
\begin{bmatrix}
\overline{\zeta}_u & \overline{\zeta}_d & \overline{\zeta}_s
\end{bmatrix}\Biggr)\overline{W}^{\mathsf{T}}
=W\left(\boldsymbol{\xi}\boldsymbol{\otimes}\overline{\boldsymbol{\zeta}}\right)\BoldExp{W}{*}=W\;\mathrm{X}\;\BoldExp{W}{*}
\nonumber
\tag{020}\label{020}
\end{align}
that is
\begin{equation}
\BoldExp{\mathrm{X}}{'} = W\;\mathrm{X}\;\BoldExp{W}{*}
\tag{021}\label{021}
\end{equation}
Above equation \eqref{021} is the transformation law of meson states in the 9-dimensional space $\;\mathbf{M}=\mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\;$ induced by a unitary
transformation $\;W \in SU(3)\;$ in the 3-dimensional space of quarks $\mathbf{Q}$.
Under this transformation law the inner product of two meson states is invariant because its relation with the trace, equation \eqref{016}, yields
\begin{equation}
\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\mathrm{Tr}\left[\BoldExp{\mathrm{X}}{'}\BoldExp{\BoldExp{\mathrm{Y}}{'}}{*}\right]=\mathrm{Tr}\Bigl[\left(W\mathrm{X}\BoldExp{W}{*}\right) \BoldExp{\left(W\mathrm{Y}\BoldExp{W}{*}\right)}{*}\Bigr]=\mathrm{Tr}\Bigl[W \left( \mathrm{X}\BoldExp{Y}{*}\right)\BoldExp{W}{*}\Bigr]=\mathrm{Tr}\Bigl[\mathrm{X}\BoldExp{Y}{*}\Bigr]
\tag{022}\label{022}
\end{equation}
The last equality in above equation \eqref{022} is valid since under the transformation law \eqref{021} the trace remains invariant. More generally, for unitary $\;W \in SU(n)\;$ and $\;A\;$ a $\;n \times n\;$ complex matrix the transformation
\begin{equation}
\BoldExp{\mathrm{A}}{'} = W\;\mathrm{A}\;\BoldExp{W}{*}
\tag{023}\label{023}
\end{equation}
if expressed in terms of elements, yields (we use the Einstein summation convention)
\begin{equation}
\BoldExp{a_{ij}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma j}}{*}
\tag{024}\label{0242}
\end{equation}
so
\begin{equation}
\mathrm{Tr}\Bigl[\BoldExp{\mathrm{A}}{'}\Bigr]=\BoldExp{a_{ii}}{'} = w_{i\rho}a_{\rho\sigma}\BoldExp{w_{\sigma i}}{*}=(\BoldExp{w_{\sigma i}}{*}w_{i\rho})a_{\rho\sigma}=\delta_{\sigma\rho}a_{\rho\sigma}=a_{\rho\rho}=\mathrm{Tr}\Bigl[A\Bigr]
\tag{025}\label{025}
\end{equation}
proving the invariance of inner product under the transformation law \eqref{021}
\begin{equation}
\langle \BoldExp{\mathrm{X}}{'},\BoldExp{\mathrm{Y}}{'}\rangle =\langle \mathrm{X},\mathrm{Y}\rangle
\tag{026}\label{026}
\end{equation}
Now, obviously the meson state represented by the identity matrix
\begin{equation}
\mathrm{I}=
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
\tag{027}\label{027}
\end{equation}
is unchanged under the transformation \eqref{021} and if normalized yields
\begin{equation}
\BoldSub{\mathrm{F}}{0}=\sqrt{\tfrac{1}{3}}
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
=\sqrt{\tfrac{1}{3}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}+\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \BoldExp{\boldsymbol{\eta}}{\prime}
\tag{028}\label{028}
\end{equation}
that is, it represents the $\;\BoldExp{\boldsymbol{\eta}}{\prime}\;$ meson.
The 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ spanned by this state is invariant. Note that $\;\BoldExp{\boldsymbol{\eta}}{\prime}=\sqrt{3}\cdot \mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]$.
Any meson state orthogonal to this space, $\mathrm{X}\perp\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace} $, remains orthogonal under the transformation. But
\begin{equation}
\mathrm{X}\perp \boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\Leftrightarrow\langle \mathrm{X},\BoldSub{\mathrm{F}}{0}\rangle =0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\BoldSub{\mathrm{F}}{0}^{\boldsymbol{*}}\right]=0\Leftrightarrow\mathrm{Tr}\left[\mathrm{X}\right]=0
\tag{029}\label{029}
\end{equation}
So, the 8-dimensional linear subspace of all meson states with traceless matrix representation is the orthogonal complement of the 1-dimensional subspace $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}\;$ and if $\;\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}\;$ is any basis which spans this space then
\begin{equation}
\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\boldsymbol{\rbrace}=\boldsymbol{\lbrace}\BoldSub{\mathrm{F}}{0}\boldsymbol{\rbrace}^{\boldsymbol{\perp}}=\Bigl\{ \mathrm{X} \in \mathbf{Q}\boldsymbol{\otimes}\overline{\mathbf{Q}}\; :\; \mathrm{Tr}\left[X\right]=0 \; \Bigr\}
\tag{030}\label{030}
\end{equation}
This space is invariant under the transformation \eqref{021}.
There are arbitrary many choices of the basis $\;\left(\BoldSub{\mathrm{F}}{1},\BoldSub{\mathrm{F}}{2},\cdots,\BoldSub{\mathrm{F}}{8}\right)\;$ but a proper one must correspond to mesons in the real world and be orthonormal if possible.
So, the normalized traceless meson state
\begin{equation}
\BoldSub{\mathrm{F}}{3}=\sqrt{\tfrac{1}{2}}
\begin{bmatrix}
1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\
0 & \boldsymbol{-}1 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}} 0 & \hphantom{\boldsymbol{-}}0
\end{bmatrix}
=\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\equiv \BoldExp{\boldsymbol{\pi}}{0}
\tag{031}\label{031}
\end{equation}
represents of course the $\;\BoldExp{\boldsymbol{\pi}}{0}\;$ meson (pion).
The basis $\mathcal{F}_{\mathbf{M}}$ may be expressed symbolically as sum of a diagonal and a traceless component
\begin{equation}
\begin{split}
&\mathcal{F}_{\mathbf{M}}=\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}+\Bigl[\mathcal{F}_{\mathbf{M}}-\Bigl(\tfrac{1}{3}\mathrm{Tr}\left[\mathcal{F}_{\mathbf{M}}\right]\Bigr)\mathcal{I}\Bigr]\\
&=\begin{bmatrix}
\dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} & \dfrac{\BoldExp{\boldsymbol{\eta}}{\prime}}{\sqrt{3}}
\end{bmatrix}
+
\begin{bmatrix}
\dfrac{\left(2\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3}{\rule[0ex]{-10pt}{0ex}} & \boldsymbol{u}\overline{\boldsymbol{d}} & \boldsymbol{u}\overline{\boldsymbol{s}}\\
\boldsymbol{d}\overline{\boldsymbol{u}} & \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}+2\boldsymbol{d}\overline{\boldsymbol{d}}-\boldsymbol{s}\overline{\boldsymbol{s}}\right) }{3} & \boldsymbol{d}\overline{\boldsymbol{s}} \\
\boldsymbol{s}\overline{\boldsymbol{u}} & \boldsymbol{s}\overline{\boldsymbol{d}} & {\rule[-2ex]{-10pt}{6ex}} \dfrac{\left(-\boldsymbol{u}\overline{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}}+2\boldsymbol{s}\overline{\boldsymbol{s}}\right)}{3}
\end{bmatrix}
\end{split}
\tag{032}\label{032}
\end{equation}
The 3rd diagonal element of the traceless component of $\mathcal{F}_{\mathbf{M}}$, if opposed and normalized, yields
\begin{equation}
\BoldSub{\mathrm{F}}{8}=\sqrt{\tfrac{1}{6}}
\begin{bmatrix}
1 & \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}}1 & \hphantom{\boldsymbol{-}}0\\
0 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}2
\end{bmatrix}
=\sqrt{\tfrac{1}{6}}\left(\boldsymbol{u}\overline{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}}-2\boldsymbol{s}\overline{\boldsymbol{s}} \right)\equiv \boldsymbol{\eta}
\tag{033}\label{033}
\end{equation}
that is, it represents the $\;\boldsymbol{\eta}\;$ meson.
(to be continued in $\boldsymbol{\S\:}\textbf{B}$)
Best Answer
The reason the signs are flipped from what you expect has to do with the fact that the antiquark transforms in the opposite way under isospin rotations. If the ordinary quark doublet is a column vector $$q=(u, d)^T$$ and transforms under rotations as $$q\rightarrow U(R) q$$ the antiquark doublet is a row vector $$\bar{q}=(\bar{u}, \bar{d})\rightarrow \bar{q} U(R)^\dagger.$$
But $SU(2)$ has a special property called being "pseudoreal" so we can write the antiquarks as a column vector that transforms normally like $$(-\bar{d}, \bar{u})^T\rightarrow U(R)(-\bar{d}, \bar{u})^T$$ This is related to the Pauli matrix $\sigma_2$ being like a charge conjugation operator if you are familiar with that.
To do the addition of isospin in the ordinary way we need both quark and antiquark in the same representation, so the singlet $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ is in this case $$u\bar{u}-d(-\bar{d})$$ so we pick up a plus sign.