So Q₁ is the total charge on C₁. Q₂ is the total charge on C₂.

This charge will be due to any initial charge ( Q₁(0) = V₀₁·C₁, Q₂(0) = V₀₂·C₂ ), and the charge flowing through the capacitors because of the switch, which we can call Q. So Q₁ = Q + V₀₁·C₁, Q₂ = Q + V₀₂·C₂.

There is no reason why Q₁ and Q₂ should be the same - for example, if V₀₁ = V₀ and V₀₂ = 0, then no current will flow when the switch is closed, Q = 0, Q₁ = C₁·V₀, and Q₂ = C₂·0 = 0.

This one is set up with V₀₁ = 2V, V₀₂ = 0V and V₀ = 4V which ends up with V₁ = 3V and V₂ = 1V, so Q₁ = 3 Q₂: simulate this circuit – Schematic created using CircuitLab

$Q_1$So Q₁ is the total charge on $C_1$C₁. $Q_2$Q₂ is the total charge on $C_2$C₂.

This charge will be due to theany initial charge ( Q₁(0) = V₀₁·C₁, Q₂(0) = V₀₂·C₂ ), and the charge flowing through the capacitors because of the switch, which we can call Q. So Q₁ = Q + V₀₁·C₁, Q₂ = Q + V₀₂·C₂.

There is no reason why Q_1Q₁ and Q_2Q₂ should be the same - for example, if V_01V₀₁ = V_0V₀ and V_02V₀₂ = 0, then no current will flow when the switch is closed and $Q_1, Q = C_1 V_0$0, but $Q_2Q₁ = 0$C₁·V₀, and Q₂ = C₂·0 = 0.

$Q_1$ is the charge on $C_1$. $Q_2$ is the charge on $C_2$.

This charge will be due to the initial charge, and the charge flowing through the capacitors because of the switch.

There is no reason why Q_1 and Q_2 should be the same - for example, if V_01 = V_0 and V_02 = 0, then no current will flow when the switch is closed and $Q_1 = C_1 V_0$, but $Q_2 = 0$.