Can one measure two components of spin exactly by measuring two components of entangled, say, electrons? Of a single electron, two different components of spin can't be have simultaneously well defined values. But what if we entangle two of them and we measure, say, $(S_1)_z$ and $(S_2)_x$ simultaneously. Wouldn't you know $(S_1)_x$ and $(S_2)_z$ then at the same time too?
 A: Say you have two (distinguishable) spin-$1/2$ particles prepared in the (Bell-) state
$$ |\psi\rangle = \frac{1}{\sqrt{2}} \left(\lvert \uparrow\rangle \otimes \lvert \downarrow\rangle - \lvert \downarrow\rangle \otimes \lvert \uparrow\rangle \right)\tag{1} \quad ,$$
where $\lvert \uparrow\rangle$ and $\lvert \downarrow\rangle$ denote the spin-up and spin-down states in the $z$-direction. Let's say you've measured the $z$-component of the first particle and you obtained $+1/2$; then the state after the measurement is given by
$$ |\tilde \psi\rangle = \lvert \uparrow\rangle \otimes \lvert \downarrow \rangle  \quad .\tag{2} $$
So after this first measurement, the probability to obtain $+1/2$ for a measurement of the spin-$z$ component of the first particle is one and for the second particle $0$ (i.e., the probability to measure $-1/2$ for the second particle is $1$). Suppose you now measure the $x$-component of the second particle, and suppose you obtain $+1/2$, then the state after the measurement is
$$|\phi\rangle = \lvert \uparrow\rangle \otimes \lvert+\rangle \tag{3} \quad ,$$
where $\lvert +\rangle$ denotes the spin-up component in the $x$-direction.
To summarize: If you measure the spin-$z$ component  of the first particle and obtain $+1/2$ and if you measure the spin-$x$ component of the second particle and obtain $+1/2$, then the state after the measurements is given by $(3)$.
It is obvious now that the probability to measure $\pm 1/2$ for the spin-$x$ component of the first particle is $1/2$ and similarly for the spin-$z$ component of the second particle.
Here, we performed two local measurements, i.e. measurements on one of the subsystems each. But we can obtain the very same result with a measurement on the overall system via a joint measurement. To see this, note that since $[S_z \otimes \mathbb I, \mathbb I\otimes S_x] =0$, we can, at least formally, construct an observable$^\dagger$
$$C:=\sum\limits_{ij} c_{ij} \, P_i \otimes Q_j \quad , \tag{4} $$
where $c_{ij} \in \mathbb R$ are all different and $P_i$, $Q_j$ are orthogonal projectors defined through the following eigendecompositions:
\begin{align}
S_z &= \sum\limits_i s_i\, P_i  \tag{5}\\
S_x &= \sum\limits_j \tilde s_j \, Q_j \tag{6} \quad .
\end{align}
In particular, we have, for example
$$P_\uparrow \otimes  Q_+\, \lvert \uparrow \rangle \otimes \lvert + \rangle  = \lvert \uparrow \rangle \otimes \lvert + \rangle  \tag{7} $$
and thus
$$C\,\lvert \uparrow \rangle \otimes \lvert + \rangle = c_{\uparrow + }\, \lvert \uparrow \rangle \otimes \lvert + \rangle \quad . \tag{8}$$
We can further construct two functions, $f$ and $g$ such that $f(c_{ij})=s_i$ and $g(c_{ij}) = \tilde s_j$, leading to $S_z \otimes \mathbb I=f(C)$ and $ \mathbb I \otimes S_x=g(C)$.
Then, upon measuring $C$ in the state $(1)$, we obtain the result $c_{ij}$, from which we can read-off the $s_i$ and $\tilde s_j$. The state after the measurement is
$$ \propto P_i \otimes Q_j |\psi\rangle \quad . \tag{9} $$
Hence, by employing $(7)$ we see that a measurement of $C$ with measurement outcome $c_{\uparrow +}$ such that $s_\uparrow=+1/2=\tilde s_+$ yields the state $(3)$ as a post-measurement state.

$^\dagger$The proof of the corresponding theorem and the details of the calculations can be found in e.g. chapter $6$ of Chris J. Isham, Lectures on Quantum Theory: Mathematical and Structural Foundations. Imperial College Press.
