Proof that user public key corresponds the curve equation (secp256r1)
It is on the curve. Used the sageMath code supplied here for the NIST P-256 curve ( aka secp256r1). Test it online here
# Finite field prime
p256 = 0xFFFFFFFF00000001000000000000000000000000FFFFFFFFFFFFFFFFFFFFFFFF

# Curve parameters for the curve equation: y^2 = x^3 + a256*x +b256
a256 = p256 - 3
b256 = 0x5AC635D8AA3A93E7B3EBBD55769886BC651D06B0CC53B0F63BCE3C3E27D2604B


px = 0x7d550bc2384fd76a47b8b0871165395e4e4d5ab9cb4ee286d1c60d074d7d60ef
py = 0x8cc6dd01e747ccb8bedaae6e7fb875d036ce7e4e6231b75b93993b15202829ac

print(&quot;The x-coordinate =&quot;, px)
print(&quot;The y-coordinate =&quot;, py)
print(py^2 % p256 )
print((px^3 + a256*px + b256) % p256)

outputs:
79438795822128695252942075663504569320626776578533931102248002205388342529032
79438795822128695252942075663504569320626776578533931102248002205388342529032

Note 1: The u2f standard can be found FIDO U2F Raw Message Formats
Note 2: there is no need to take the square root. Just compare both sides in the modulus of the curve. This eliminates extra square root calculations on the field.
A little elliptic curve;
In the finite field based Elliptic Curve, we take a finite field $F$ (prime field or finite field extension) and define a curve equation by $$y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6$$ This is Weierstrass equation.
If the field $F$ has characteristics different from 2 or 3 then the Weierstrass equation can be turned into the short form.
$$y^2 = x^3 + Ax +B$$
The equation is the standard short Weierstrass form. We don't want a singular curve so we check that the curve discriminant is non zero by $$2A^3-27B^2 \neq 0.$$
The points of the curve form an abelian group under the geometric interpretation with an identity element call $\mathcal{O}$, it can be the point at infinity like in secp256r1 or $(0,1)$ as the Edwards curves. The geometric interpretation of point addition turned into arithmetic over the field $F$.