Difficulty understanding 'random square root' in enhancement to fiat-shamir
So the corresponding $v_i$ for $s_1$ would be 2. If $n=pq=11∗19=209$, then does this mean that $s_i = \sqrt{2} \bmod n$?


What they mean by "square-root of $x$" is the value $y$ that is a solution to $y^2 \equiv x \pmod n$; as there are multiple solutions (in the cases that at least one solution exists), then by a random solution, they mean to pick one randomly.

Now, in the case of $n=209$, it turns out that there isn't a solution for $v_i = 2$ or $3$.  However, there are four solutions for $v_i = 5$, namely 29, 48, 161, 180 (and for $n$ which is a product of two distinct odd primes and with $v_i$ nonzero and small (that is, $&lt; p, q$), there will always be either zero or four solutions); the algorithm is to set $s_i$ to one of the four randomly.

As for how to compute the square root of $x$, the procedure is:


Compute $x_p = x \bmod p$ and $x_q = x \bmod q$
Compute the values $y_p$ that satisfies $y_p^2 \equiv x_p \bmod p$ and the values $y_q$ that satisfies $y_q^2 \equiv x_q \bmod q$; and if there doesn't exist either a $y_p$ or a $y_q$ that satisfies the above equations, then there doesn't exist such a $y$.  This computation can be done in the general case by the Tonelli–Shanks algorithm (and with simpler algorithms in the case that $p, q \not\equiv 1 \pmod 8$)
Usually, the above will give two possible values for $y_p, y_q$; pick one randomly (this will perform the 'random selection' logic specified.
Use the Chinese remainder theorem to find the value $y$ with $y \equiv y_p \pmod p$ and $y \equiv y_q \pmod q$; you're done