It is enough to show that $𝕊^T A=B$, since then $𝕊=(B A^{-1})^T=(A^T)^{-1} B^T=A^{-1} B$, using that $A, B$ are symmetric. In local coordinates, for $v, w ∈ ℝ^2$, the first fundamental form is $I(v, w)=v^T A w$, the second is $I I(v, w)=v^T B w$. The shape operator is defined by $I(𝕊 v, w)=I I(v, w)$. The latter formula becomes $v^T 𝕊^T A w=v^T B w$. As this holds for all $v, w$, we deduce $𝕊^T A=B$.

Prove that there are exactly two local parametrizations $F(x, y)$ of $S$, defined for sufficiently small real numbers $x, y$ with $F(0,0)=p$ and $F(x, 0)=γ(x)$, such that $y ↦ F(x, y)$ are unit speed geodesics perpendicular to $γ(t)$.

[Provided you state them clearly, you may use standard existence and uniqueness results about geodesics, and criteria for when a map is a local parametrization.]

Standard result:

Note by definition that $∂_y F(x, 0)=n(x) × γ'(x)$ is a unit vector perpendicular to $γ'(x)$ (using that $|γ'(x)|=1$, since $γ$ has unit speed). The two choices of $n$ correspond to the two choices $n(x) × γ'(x)$ of a unit tangent vector field defined along $γ$ near $p$ and perpendicular to $γ$.

Geodesics have constant speed since $\frac{d}{d t} I(γ', γ')=2 I(∇_t γ', γ')=0$ (using the geodesic equation $∇_t γ'=0$). Therefore $F(x, y)$ has constant speed $1=|n(x) × γ'(x)|$.

To prove that $F(x, y)$ is a local parametrization, it suffices to show that near $p$ the derivatives $∂_x F, ∂_y F$ are linearly independent. This holds along $γ$ by construction (since they are perpendicular there), hence it holds near $γ$ by continuity (using that $G_{p, v}$ depends smoothly on $p, v$).

given $p ∈ S, v ∈ T_p S$, there is a unique geodesic $G_{p, v}:(-ε, ε) → S$, defined for small time $ε>0$, with $G_{p, v}(0)=p, G_{p, v}'(0)=v$, moreover $G_{p, v}$ varies smoothly in $p, v$.Let $F(x, y)=G_{γ(x), n(x) × γ'(x)}$ where $n(x)$ is a unit vector field normal to $S$ along $γ(x)$ defined near $p$. Note there are exactly two choices of such a normal vector field.

Note by definition that $∂_y F(x, 0)=n(x) × γ'(x)$ is a unit vector perpendicular to $γ'(x)$ (using that $|γ'(x)|=1$, since $γ$ has unit speed). The two choices of $n$ correspond to the two choices $n(x) × γ'(x)$ of a unit tangent vector field defined along $γ$ near $p$ and perpendicular to $γ$.

Geodesics have constant speed since $\frac{d}{d t} I(γ', γ')=2 I(∇_t γ', γ')=0$ (using the geodesic equation $∇_t γ'=0$). Therefore $F(x, y)$ has constant speed $1=|n(x) × γ'(x)|$.

To prove that $F(x, y)$ is a local parametrization, it suffices to show that near $p$ the derivatives $∂_x F, ∂_y F$ are linearly independent. This holds along $γ$ by construction (since they are perpendicular there), hence it holds near $γ$ by continuity (using that $G_{p, v}$ depends smoothly on $p, v$).

\[
F(θ, φ)=\begin{pmatrix}
\sin θ \cos φ \\
\sin θ \sin φ \\
\cos θ
\end{pmatrix} ∂_θ F=\begin{pmatrix}
\cos θ \cos φ \\
\cos θ \sin φ \\
-\sin θ
\end{pmatrix} ∂_{φ} F=\begin{pmatrix}
-\sin θ \sin φ \\
\sin θ \cos φ \\
0
\end{pmatrix}
\]
Thus
\[
I=∂_θ F ⋅ ∂_θ F d θ^2+2 ∂_θ F ⋅ ∂_{φ} F d θ d φ+∂_{φ} F ⋅ ∂_{φ} F d φ^2=d θ^2+\sin ^2 θ d φ^2
\]
The geodesics on the sphere are great circles. The equator is $θ=π / 2$ so it is defined by the local equations $θ=π / 2, φ=t$. This is a unit speed parametrization since: $θ'=0, φ'=1$ and so $I(γ', γ')=(θ')^2+\sin ^2 θ(φ')^2=1$ (so unit speed).

A geodesic $y ↦ F(x, y)$ orthogonal to the equator at $φ=x$ is the great circle which goes through the North/South poles given by $θ=y, φ=x$(since $(\sin y \cos x, \sin y \sin x) ∈ ℝ^2$ is the line segment in the plane with slope $\tan x$ written in polar coordinates, and $z=\cos y$ varies between -1 and 1 which are the two Poles). Since $∂_y F$ corresponds to $θ'=1, φ'=0$ we again see from $I$ that $y ↦ F(x, y)$ has unit speed. Thus the two possible parametrizations from Part (b) are $x=φ, y=θ$ or $x=φ$, $y=-θ$. [If we instead picked the equator to run in the opposite direction, then the sign of $φ$ would be reversed everywhere]

A geodesic $y ↦ F(x, y)$ orthogonal to the equator at $φ=x$ is the great circle which goes through the North/South poles given by $θ=y, φ=x$(since $(\sin y \cos x, \sin y \sin x) ∈ ℝ^2$ is the line segment in the plane with slope $\tan x$ written in polar coordinates, and $z=\cos y$ varies between -1 and 1 which are the two Poles). Since $∂_y F$ corresponds to $θ'=1, φ'=0$ we again see from $I$ that $y ↦ F(x, y)$ has unit speed. Thus the two possible parametrizations from Part (b) are $x=φ, y=θ$ or $x=φ$, $y=-θ$. [If we instead picked the equator to run in the opposite direction, then the sign of $φ$ would be reversed everywhere]

Let the given curve $c(t)$ be described locally by $(θ(t), φ(t))$. The tangent to the meridian line is $φ=\text{const}$ which is locally $(1,0) ∈ ℝ^2$. Thus imposing angle $α$ between the curve and the meridian gives
\[
\cos α=\frac{I(∂_{φ} F, c')}{\sqrt{I(∂_{φ} F, ∂_{φ} F)} \sqrt{I(c', c')}}=\frac{θ'}{\sqrt{(θ')^2+\sin ^2 θ(φ')^2}}
\]
Thus $(θ')^2=\cos ^2 α ⋅((θ')^2+\sin ^2 θ(φ')^2)$, so $(θ')^2 \sin ^2 α=\cos ^2 α \sin ^2 θ(φ')^2$, thus
\[
\frac{θ'}{\sin θ}= ± \frac{φ'}{\tan α}
\]
where $±$ arises from taking the square root. Integrating (using $\frac{d}{d x} \log \tan \left(\frac{x}{2}\right)=\frac{1}{\sin x}$) gives the result, where $c$ arises because of the constant of integration.