Proof of SAT is randomly reducible to UNIQUE-SAT
Q: $x_1,...x_n$ are unknown literals, how can we construct $f_k$ with unknown literals and where $y_1,...,y_m$ came from?


$x_1, \ldots, x_n$ are variables, each of which can appear as a positive literal ($x_i$) or a negative one ($\lnot x_i$). To construct an equivalent formula $f_1$ of another formula $f_1'$, you don't have to know the values (truth assignments) of $f_1'$. (A simple example is $x_1 \to x_2 \Leftrightarrow \lnot x_1 \lor x_2$.) 

The variables $y_1, \ldots, y_m$ are unimportant (at least in this Lemma). The Lemma define the "equivalence" of the two formulas $f_1$ and $f_1'$ by equality on the $x_i$ values (leaving $y_i$ values unspecified). 




  Q: How do we come to the such construction of $f'_1$ and why showing only $f'_1$ sufficient for the proof?


Each $f_i$ (and $f_i'$) is independent and similar. Therefore, the case of $f_1'$ is sufficient for the proof. To justify the construction of $f_1' = f \land (x_{i_1} \oplus x_{i_2} \oplus \cdots \oplus x_{i_j} \oplus 1)$, you should prove the following two points:


Each solution $v$ to $f_1'$ satisfies $f$.


This is trivial by the definition of $f_1'$.


Each solution $v$ satisfies the equation $v \cdot w_1 = 0$. 


This is the tricky part. We argue as follows:

First, $$v \cdot w_1 = x_{i_1} + x_{i_2} + \cdots + x_{i_j} \text{ (here `+` is over GF[2])} = x_{i_1} \oplus x_{i_2} \oplus \cdots \oplus x_{i_j}$$ 
The first equality follows from the fact that other $w_l$ with indices $l \neq i_1, i_2, \cdots, i_j$ are all 0 (according to the definition of all $x_{i_j}$).

Because $v$ is a solution to $f_1'$, we have $$x_{i_1} \oplus x_{i_2} \oplus \cdots \oplus x_{i_j} \oplus 1 = 1$$
Thus,
$$v \cdot w_1 = x_{i_1} \oplus x_{i_2} \oplus \cdots \oplus x_{i_j} = 0$$