Question about the Simon's algorithm
I'm not sure I get all your questions. I'll try to answer according to my interpretation of what you are asking, and you can clarify if I got you wrong.


  
  Given this I want to understand why a measurement on the second qubit would necessarily collapse this to the state, $\psi_b = \frac{1}{\sqrt{2}} ( |j\rangle + |j \oplus s\rangle)|x_j\rangle$ for some $j \in \{0,1\}^n$? 
  


First, there is no second qubit, but a second system (of $n$ qubits). Next, consider $\psi_a$, and recall that for any $f(x_j)$ there are only two preimages $j$ and $j\oplus s$, then if you measure $f(x_j)$ at the second system, the first system must either be $j$ or $j\oplus s$.


  
  Is this measurement of the second qubit state trying to measure some observable/Hermitian operator whose eigenstates are precisely these $\psi_b$s? Then it would make some sense that the only things one sees are these $\psi_b$ states.  
  


Kind of. However, you don't measure $\vert\psi_a\rangle$ directly. There is another hadamard gate which is missing there. I.e., you measure
$$(H\otimes I) \left\vert\psi_a\right\rangle = \frac1{2^n}\sum_x\sum_i (-1)^{i\cdot x}\left\vert i\right\rangle\left\vert f(x)\right\rangle$$

Now if you re-arrange it properly, it should give
\begin{align*}
\frac1{2^n}\sum_{x\in\{0,1\}^n}\sum_i (-1)^{i\cdot x}\left\vert i\right\rangle\left\vert f(x)\right\rangle &amp;=
\frac1{2^n}\sum_{j\in XS}\sum_i (-1)^{i\cdot j}\left\vert i\right\rangle\left\vert f(j)\right\rangle + (-1)^{i\cdot (j\oplus s)}\left\vert f(j\oplus s)\right\rangle \\
&amp;=
 \frac1{2^n}\sum_{j\in XS}\sum_i \left ((-1)^{i\cdot j}+(-1)^{i\cdot j}(-1)^{i\cdot s}\right)\left\vert i\right\rangle\left\vert f(j)\right\rangle \\
&amp;=
 \frac1{2^n}\sum_{j\in XS}\sum_i (-1)^{i\cdot j}\left(1+(-1)^{i\cdot s}\right)\left\vert i\right\rangle\left\vert f(j)\right\rangle 
\end{align*}
Where $XS\subseteq \{0,1\}^n$ is the largest subset such that if $j\in XS$ then $\forall j'\in XS, f(j)\ne f(j')$; so that the sum over $j$ is only over half the space and  we don't count twice $x=j$ and $x=j\oplus s$. But note that $f(j)=f(j\oplus s)$ which we use in the second transition.

The above shows that for any $i$ for which $s\cdot i=1 \mod 2$ there is a destructive interference that cancels these terms, and you are left with only $i$'s for which $s\cdot i=0 \mod 2$. Thus, any $i$ you will measure (at the first system) must satisfy $s\cdot i=0 \mod 2$.