Secret Sharing between $x$ parties So for an unknown number of parties $x$, I make $x-1$ random $n$-bit strings represented by {$r_1, r_2,...,r_{x-1}$}. Where  $s$ is the complete string split up between $x$ parties (yes?). 

I then send each party one of the fragments $s_i$, so party 1 would get $s_1$, party 2 would get $s_2$ etc. all the way through to $s_{x-1}$. Along with these fragments. I then send $r_1⊕...⊕r_{x-1}⊕s$ to party $x$.

The message $s$ can then be reconstructed by XORing all the fragments together. Any singular $x-1$ party should not have enough information to reconstruct a message. But then what happens if there are only two parties? if $x=2$, then the message $s$ would be divided between $x-1=1$, which means that that singular party would acquire the entire $n$-bit string on their own, or am I over looking something that prevents this so that any $x-1$ party does infact not have enough information to reconstruct the message, even if it was only 1-bit ($n=1$) in length?

What about if 3 parties share two keys $a,b$? Assuming their fragments are $f_j$ for $a$ and $g_j$ for $b$ (that is these are $r_j$ fragments, and party 1 has $g_1$ and $f_1$, party 2 has  $g_2$ and $f_2$ etc), could they manage to derive the secret if each party makes $z_j=f_j⊕g_j$, using the $z_j$ values of each?
So for an unknown number of parties $x$, I make $x-1$ random $n$-bit strings represented by {$r_1, r_2,...,r_{x-1}$}. Where  $s$ is the complete string split up between $x$ parties (yes?). 
  
  I then send each party one of the fragments $s_i$, so party 1 would get $s_1$, party 2 would get $s_2$ etc. all the way through to $s_{x-1}$. Along with these fragments. I then send $r_1⊕...⊕r_{x-1}⊕s$ to party $x$.


That is not how it is typically done.  One issue is that two parties (say, party 1 and party $x$) get partial information of the shared secret, that is, the part that was masked by $r_1$.

Instead, what we do is make $x-1$ random strings (each as long as the secret $s$).  Then, we send party 1 the string $r_1$ (not $s_1$), party 2 would get $r_2$, etc.  And then send $r_1 \oplus r_2 \oplus ... \oplus r_{x-1} \oplus s$ to party $x$.

This can be seen to yield no information about $s$ to any $x-1$ parties, and still makes $s$ recoverable to someone with all $x$ shares.

Studying this version of the protocol may answer your questions.