Let me add to the other answers with a more concrete iteration.
With precision I mean the number of bits used per $2$-adic integer.
Hensel lifting resembles Newton iteration.
The usual Newton-Raphson scheme for the reciprocal squareroot
also works for $p$-adic squares, provided you begin with a close-enough
initial guess,
which here means that the initial unit digit must be correct.
Multiplying $a$ with its reciprocal squareroot $1/\sqrt{a}$ gives you the
ordinary squareroot $\sqrt{a}$.
Newton-Raphson computation of $x = \frac{1}{\sqrt{a}}$ finds a zero of
$f(x)=\frac{1}{a x^2}-1$ using the iteration
$$x_{n+1} = x_n\,(3 - a x_n^2)/2,$$
provided that $a\equiv1\pmod{8}$ and $x_0$ is odd. The next bit (weight $2$)
of $x_0$ is preserved by the iteration; think of it as deciding on the sign
of the squareroot to return. So basically you begin with two correct bits.
From there on, each step first doubles the number of correct bits,
then loses one bit due to the division by $2$.
A note on the division by $2$. No problem:
Division by $2$ is defined in $\mathbb{Q}_2$, and it yields
a $2$-adic integer if the dividend is an even $2$-adic integer.
This is the case here, as $a$ and all the $x_n$ are odd.
So just shift down 1 bit.
However, when working with fixed finite precision, this means that something
needs to get shifted into the highest bit. The correct value would depend
on $a$'s next higher bit which you do not know, but either choice works
in the sense that squaring with the same precision yields the same result.
This is why there are four possible solutions with finite precision.
If you consider that highest bit an inaccuracy and remove it from the result,
there are only two possible solutions, depending on your choice of
$x_0\equiv\pm1\pmod{4}$.