Starting with the state $|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} $.
If you want to find the probability of measuring $+1$ in observable $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $ for the first qubit, and $+1$ in the observable $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $ for the second qubit then you can calculate it as $\langle \psi| M | \psi \rangle = Tr(\rho M)$ where $\rho = |\psi \rangle \langle \psi |$ and here $M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$ since $|+\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $X$ and $|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ is the eigenvector corresponding to the $+1$ eigenvalue of $Z$. So explicitly
\begin{align} M = |+\rangle \langle+| \otimes |0\rangle\langle 0| &= \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \bigg[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \bigg] \\
&= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \\
&= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}
\end{align}
Thus,
$$\langle \psi| M | \psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 1 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \dfrac{1}{4} $$
Also note that, the above is the same if we have done $Tr(\rho M)$ since
$$Tr\bigg( \rho M \bigg) = Tr\bigg( \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \bigg) = \dfrac{1}{4}$$
You can extend this to other cases as well.
Update:
If you want to do sequential measurement, then you can find the the post measurement state $|\psi\rangle_{post}$ then follow the same procedure.
For instance, if we again start with $|\psi \rangle = \dfrac{|01\rangle + |10 \rangle }{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} $
and we want to find the probability of measuring $+1$ in observable $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $ for the first qubit. Then afterward, finding the probability of measuring $+1$ in observable $X $ for the first qubit, and $+1$ in the observable $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $ for the second qubit on this collapsed state then we can do it as follow:
First Step: To find the probability of measuring $+1$ in observable $X $ we can construct $M $ as
\begin{align} M = |+\rangle \langle+| \otimes I = \bigg[ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \bigg] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} &= \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \\
&= \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix}
\end{align}
And therefore,
$$\langle \psi| M | \psi \rangle = \dfrac{1}{2} $$
and the state after measurement, $|\psi_{post}\rangle $, is going to be
\begin{align}
|\psi_{post}\rangle = \dfrac{ M |\psi \rangle }{ \sqrt{prob(+1)}} = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix} \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} }{ \sqrt{ 1/\sqrt{2} } } = \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}
\end{align}
Second Step: Now the probability of measuring $+1$ in observable $X $ for the first qubit, and $+1$ in the observable $Z $ for the second qubit on this collapsed state $|\psi_{post} \rangle$ can be calculated as $\langle \psi_{post} | M | \psi_{post} \rangle $ where again $M = |+\rangle\langle +| \otimes |0\rangle\langle 0 |$ (as indicated why on the top of this answer). Hence this probability is
$$
\langle \psi_{post} | M | \psi_{post} \rangle = \dfrac{1}{2} \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix} \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = \dfrac{1}{2}
$$
Where the post state after this process, $|\psi_{post 2} \rangle$ is now in the state
$$
|\psi_{post 2} \rangle = \dfrac{ \dfrac{1}{2}\begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \dfrac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} }{ \sqrt{1/2} } = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \dfrac{|00\rangle + |10\rangle }{\sqrt{2}}
$$
