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 $𝕊^T A=B$.

Standard result:

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).
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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]
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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.