Working out the example from your comment. Here $p=127$, and you want to compute modular square roots of $a=13290059$. First we notice that $a\equiv 17 \pmod p$ and $a\equiv 15892 \pmod {p^2}$. You have apparently found that $12^2\equiv 17 \pmod p$, so the two square roots will be congruent to $\pm12\pmod p$. I do the plus sign here.
We are looking for an integer $b$ such that $b^2\equiv 15892 \pmod{127^2}$ and $b\equiv 12 \pmod {127}$. The latter congruence tells us that $b=12+127 m$ for some integer $m$. Then
$$
b^2=12^2+24\cdot 127 m+127^2m^2\equiv 144+24\cdot127 m\pmod{127^2}.
$$
Here $15892-144=15748=127\cdot124$, so the first congruence simplifies first to
$$144+24\cdot127 m\equiv 15892\pmod{127^2}\Leftrightarrow 24\cdot127 m\equiv 127\cdot124\pmod{127^2},$$
and then by cancelling a factor $127$ to
$$
24m\equiv124\pmod{127}.
$$
The inverse of $24$ modulo $127$ is $90$, so the unique solution of this is
$$
m\equiv 90\cdot124\equiv 111\pmod{127}.
$$
This corresponds to the square root
$$
b=12+111\cdot127=14109.
$$
There are actually several different concepts here, so I'll try to address all of them. I'll get to the modular arithmetic in just a moment, but first a review:
SQUARE ROOTS
We should know that 25 has two square roots in ordinary arithmetic: 5 and -5.
MODULAR ARITHMETIC SQUARE ROOTS
IF the square root exists, there are 2 of them modulo a prime. To continue our example, 25 has the two square roots 5 and -5.
We can check this:
$$(-5)^2 = 25 \equiv 3\bmod 11$$
$$(5)^2 = 25 \equiv 3\bmod 11$$
To find the square roots sometimes takes a bit of trial and error. Often you have to go through each value $v$ and square it (to get $v^2$) to check if it's equivalent to $n \bmod p$, where $n$ is the number whose square root you want to find.
MULTIPLE PRIMES
Again, if a square root exists, there are two square roots modulo each prime. So if we are using multiple primes, there can be more square roots. For example, with two primes, there are 2 square roots modulo the first prime and two square roots modulo the second prime. This gives us $2 \cdot 2 = 4$ square roots.
In general, if we can find a square root modulo each prime, there are a total of $2^n$ square roots modulo $n$ primes.
RETURNING TO YOUR EXAMPLE
We can first calculate 3 modulo 11 and 13:
$$3 \equiv 3 (\bmod 11)$$
$$3 \equiv 3 (\bmod 13)$$
So, modulo 11, we are looking to find a number that, when squared, is equivalent to 3. If we find one, we know that there will be another. So we check the numbers: $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 9, \dots$ and find that
$$5^2 = 25 \equiv 3 (\bmod 11)$$
...so we know that there will also be another square root modulo 11. Continuing on our quest, we check
$$6^2 = 36 \equiv 3 (\bmod 11)$$
...so we've found the square roots modulo 11. We then continue this modulo 13 to find that:
$$4^2 = 16 \equiv 3 (\bmod 13)$$
$$9^2 = 81 \equiv 3 (\bmod 13)$$
So we know that our square root is either 5 or 6 modulo 11, and either 4 or 9 modulo 13. This gives us 4 possibilities. We can then find that:
$$82 \equiv 5 (\bmod 11), 82 \equiv 4 (\bmod 13)$$
$$126 \equiv 5 (\bmod 11), 126 \equiv 9 (\bmod 13)$$
$$17 \equiv 6 (\bmod 11), 17 \equiv 4 (\bmod 13)$$
$$61 \equiv 6 (\bmod 11), 61 \equiv 9 (\bmod 13)$$
Best Answer
An algorithm to compute $\sqrt{c} \pmod {p^2}$ from $\sqrt{c} \pmod {p}$ is given below. It is a simplified version of Alg 2.3.11 (Hensel lifting) from Crandall/Pomerance: Prime Numbers, A Computational Perspective, 2nd ed., 2005 using their notation with $f(x)=c-x^2,\,$ $f'(x)=-2x.$ I include the values for $p=17, c=13$ with the root $r\equiv \sqrt{13} \equiv 8 \pmod {17}$
Check: $59^2 \equiv 3481 \equiv 13 \pmod {289}$
And with some larger values using $p=10000000019, c=5:$
See also the section Powers of odd primes of John Cook's Solving quadratic congruences and the already cited Wikipedia section from my comment (using $p=7, c=2$).