The topic - the general form of neutral current interactions
Suppose the $SU_{L}(2)\times U_{Y}(1)$ electroweak interactions. With corresponding gauge couplings $g_{1}, g_{2}$ and fixed quarks doublets $Q_{i}$,
$$
\tag 0 Q_{1} = \begin{pmatrix} u\\ d \end{pmatrix}, \quad Q_{2} = \begin{pmatrix}c \\ s\end{pmatrix}, \quad Q_{3} = \begin{pmatrix}t \\ b \end{pmatrix}
$$
and the quarks column $q_{a}$,
$$
\tag 1 q_{a} = (u,c,t,d,s,b)^{T},
$$
The quarks ''names'' are fixed by the diagonalization of the Higgs interaction term (i.e., the mass term).
You may write down the most general neutral currents interaction lagrangian
$$
\tag 2 L_{\text{int}} = g_{1}\bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c.
$$
Here
- $a_{ij}$ is the $3\times 3$ unitary matrix acting in a 3-dimensional space of quarks doublets $Q_{i}$ $(0)$,
- $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-dimensional space of quarks $q_{a}$ $(1)$,
- $\sigma_{3} = \text{diag}(1,-1)$ is the isospin zero generator of the $SU_{L}(2)$ group,
- $\theta$ denotes the Weinberg angle relating the couplings $g_{1},g_{2}$ to the EM interaction constant $\alpha$. $c(\theta)$ denotes $\cos(\theta)$, and so on.
In general, the forms of $a_{ij}$ and $c_{ab}$ are restricted. Since You already know the electric charges of quarks (precisely, the charges of an upper quarks are $\frac{2}{3}$, while the charges of the lower quarks are $-\frac{1}{3}$), the matrix $c_{ab}$ has the form
$$
\tag 3 c_{ab} = \begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix},
$$
with $A,B$ being the unitary $3\times 3$ matrices. ALso, with precise values of electric charges, You have to impose some constrains on diagonal elements of $c_{ab}, a_{ij}$, but we're not interested in this at the moment.
From $(2)$ You can extract the $Z$-boson interaction part:
$$
\tag 4 L_{\text{int}} = \bar{q}_{a}\gamma^{\mu}(c^{ab}_{V} - c^{ab}_{A}\gamma_{5})q_{b}Z_{\mu} + h.c.,
$$
with $c^{ab}_{V,A}$ having the general form of $(3)$. For example, in the Standard model we have $c^{ab}_{V,A} = \delta^{ab}c^{b}_{V,A}$ (see below), which means that there is assumed that the neutral current interactions are quark species diagonal.
Experimental origin - no so much oscillations
So why we choose
$$
\tag 5 c^{ab}_{V,A} = \delta^{ab}c^{b}_{V,A}?
$$
The answer is the experiment.
Before I'll discuss the case of quarks, let's establish the difference between the quarks and leptons cases. In the leptons case, You also may write down the matrices $c_{A,V}^{ab}$ in the form $(3)$. However, there is experimental fact to write them in the form of $(4)$ - the conservation of the so-called lepton numbers. There are 3 different lepton numbers, and this immediately imposes $(5)$. In the case of quarks such argument isn't valid. Unlike the case of leptons, there is only one conserved quark global number - the so-called baryon number (all of quarks are massive, unlike the leptons, and this kills 2 "missing" numbers). All quarks have the same baryon number, and the restriction to conserve it doesn't reduce the general form $(3)$ of $c^{ab}_{V,A}$ to a simpler one.
Let's now go to the reason why $c^{ab}$s are reduced to $(5)$ in the case of quarks. Suppose the neutral mesons $M^{0}$ - the bounded states of two quarks with zero total electric charge. They are
$$
M^{0} = \{ B^{0} = d\bar{b}, \quad \bar{B}^{0} = \bar{d}b, \quad D^{0} = c\bar{u}, \quad \bar{D}^{0} = \bar{c}u, \quad K^{0} = d\bar{s}, \quad \bar{K}^{0} = \bar{d}s \}
$$
With the general form $(3)$ of the matrices $c^{ab}_{V,A}$ there are tree-level oscillations
$$
M^{0} \to M^{0}
$$
There are also tree-level decays on a lepton-antilepton pair $l\bar{l}$,
$$
M^{0} \to l \bar{l},
$$
if we take into account the lepton part of $(4)$. Such processes violate some quantum numbers like strangeness.
Apart from these tree-level processes, there are also corresponding loop-mediated processes. In compare to the formers, the latters are suppressed by the factor of $\frac{m_{q}^{2}}{m_{Z}^{2}}<<1$, where $m_{q}$ is the mass of given quark, while $m_{z}$ is the mass of $Z$-boson. However, they are possible even in the case of $(5)$.
The experiment says that the above processes are strongly suppressed. This requires to set $c^{ab}_{V,A}$ to the form $(5)$.
