About the familiar formula $Q=I_3 + Y/2$. As an example, the left-handed electron has charge $Q=-1$, weak isospin $I_3 = -1/2$, and weak hypercharge $Y=-1$.
The electron is represented in the Dirac equation by bispinors. For example, Peskin and Schroeder's classic textbook "An Introduction to Quantum Field Theory" uses the Weyl or chiral representation where (see P&S equation 3.25): \begin{equation} \gamma^0 = \left(\begin{array}{cccc}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{array}\right). \end{equation} The charge operator $Q$ is proportional to $\gamma^0$, so that, for example, particles are eigenstates of this operator with eigenvalue $+1$ while the antiparticles take eigenvalue $-1$. See P&S equation 3.47 for the particle (positive frequency) plane wave solution to the Dirac equation.
On the other hand, the operator for handedness in this representation is (see P&S equation 3.72): \begin{equation} \gamma^5 = \left(\begin{array}{cccc}-1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right). \end{equation} and this anticommutes with $\gamma^0$.
In choosing a "complete of commuting observables" for the Dirac spinor we have the choice of "spin and charge" or "spin and handedness". We cannot use "charge and handedness" as these correspond to operators that do not commute.
P&S use the example of a spin-up and a spin-down state and give them a large boost to obtain (approximately) a pure right handed and pure left handed state in equations 3.52 and 3.53. You can see that before boosting they are eigenstates of charge and afterwards are eigenstates of $\gamma^5$, i.e. P&S equation 3.53: \begin{equation} u(p) = \left(\begin{array}{c} \sqrt{E-p^3\left(\begin{array}{c}0\\1\end{array}\right)}\\ \sqrt{E+p^3\left(\begin{array}{c}0\\1\end{array}\right)} \end{array}\right) \rightarrow \sqrt{2E}\left(\begin{array}{c}0\\1\\0\\0\end{array}\right). \end{equation} The above shows a spin-down electron being boosted to a left-handed electron.
Now boosting itself does not change the charge. But when the particle is no longer stationary it is no longer an eigenstate of charge. If you work it out you find out that while the (small) boosted states still have charge -1 in the sense that the average of charge measurement is $<Q> = -1$, a single measurement of charge will give numbers that go to $\pm \infty$. And of course using the pure left handed state on the right hand side of the above will give $<Q>=0$.
Another way of explaining the difficulty is to replace the spin-down electron with a spin-down positron. The result of a "large" boost is the same as the positron: \begin{equation} \bar{u}(p) = \left(\begin{array}{c} \sqrt{E-p^3\left(\begin{array}{c}0\\1\end{array}\right)}\\ -\sqrt{E+p^3\left(\begin{array}{c}0\\1\end{array}\right)} \end{array}\right) \rightarrow \sqrt{2E}\left(\begin{array}{c}0\\1\\0\\0\end{array}\right). \end{equation} In short, there are no states that have good quantum numbers for both charge and handedness so there are no states that have good quantum numbers for all three of charge, weak hypercharge and weak isospin. Then why can we find tables for these three quantum numbers?
Now this is essentially a quantum mechanical analysis of the relationship between charge, weak hypercharge and weak isospin. Is it that this problem goes away in a field theory analysis? And what exactly is the reason for the difference?