Let $x_1,x_2$ be Riemannian normal coordinates centered in $p\in M=S^2$, the 2-sphere, and $g\in T^*M\otimes T^*M$ the riemannian metric.
I have encountered the following statement, which I'm trying to grasp:
We have $g_{11}(0,y)=cos^2(y)$ along $(0,x_2)$.
Here, $g_{ij}:= g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})$ for the induced $T_pM$-basis $\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^2}$.
I can visualize $y\mapsto (0,y)$ as a great circle, but I'm trying to approach this rigorously from a differential geometry viewpoint.
What I understand: $(0,y)$ is really the image of a point $q\in U_p$ under the chart $(x,U_p)=(\pi\circ exp^{-1}_p,U_p)$, where $\pi:T_pM\rightarrow \mathbb{R}^2$ is the canonical isomorphism $\pi(\frac{\partial}{\partial x^i})=x_i$. So,
$$ \begin{align*} &x(q)=\pi(exp^{-1}_p(q))=(0,y)\hspace{20pt}&\in\mathbb{R}^2\\ \Leftrightarrow & exp^{-1}_p(q)=\pi^{-1}(0,y)=y\frac{\partial}{\partial x^2} &\in T_pM\\ \Leftrightarrow & q=exp_p(y\frac{\partial}{\partial x^2}) &\in U_p \end{align*} $$
And, by definition of the exponential map, $exp_p(y\frac{\partial}{\partial x^2})=\gamma_{(y\frac{\partial}{\partial x^2})}(1)=\gamma_{(\frac{\partial}{\partial x^2})}(y)$ for the geodesic $\gamma$ starting in $p$ with the velocity field written in the index.
Question: How can I, from this information (as far as it is correct, please correct anything that isn't) about the normal coordinates, conclude the form of the metric in $q$ under $(x,U_p)$, in particular the cited statement above?
I do know from the orthonormality of normal coordinates that $g_{ij}=\delta_{ij}$ in $p$, but do I know anything about how $g$ is parallel transported along $\gamma_{\frac{\partial}{\partial x^2}}$?
I think your writing is correct, but it did not start to address the question yet. As you are asking, you have to start calculating $g_{11}(0, y)$.
Concretely, that means you have to compute the image of $\frac{\partial}{\partial x^1}$ at point $(0, y)$ under the exponential map and compute its norm squared on the sphere. Formally, $$ g_{11}(0, y) = \Big\| d\exp_p\big|_{(0, y)} \big(\frac{\partial}{\partial x^1}\big)\Big\|^2, $$ where the norm squared is computed in the metric on $S^2$.
Traditionally, such push-forward under the exponential map at a different point is computed using the Jacobi fields along your geodesic $\gamma$.
In this case, you can also just choose a curve $c(t)$ in your tangent plane such that $$ c(0) = (0, y),\ \dot c(0)=\frac{\partial}{\partial x^1}=\langle 1, 0\rangle. $$
You may even choose this curve $c$ to be the circle in the tangent plane as $$ c(t) = (y\sin(t/y), y\cos(t/y)),\quad \text{say }y>0. $$ Then the image curve $$ \tilde c(t) = \exp_p c(t) $$ is a parallel circle on the sphere. In the standard spherical coordinates, we have that the image curve $\tilde c(t)$ is defined by $$ \phi = y,\ \theta = t/y $$ (or $\pi/2-t/y$ to be totally correct since $c(t)$ is not conventionally parametrized).
In any event, the length of the tangent vector of this curve (the wanted push-forward) is well known on $S^2$. It is $$ \frac{\sin y}{y}. $$ ($\sin y$ is the radius of the small circle, and $1/y$ by our parametrization.)
Therefore $$ g_{11}(0, y) = \frac{\sin^2 y}{y^2}. $$
Sorry, I didn't warn you that I did not get what you asked for, but I think my calculation is correct.
To be complete, you can use spherical coordinates $(\phi, \theta)$ to get the $S^2$ metric in geodesic normal coordinates as $$ d\phi^2 + \sin^2\phi\, d\theta^2. $$ Since $\phi$ and $\theta$ are just polar coordinates in the tangent plane, you can use $$ \phi=\sqrt{x^2+y^2},\ \theta = \arctan \frac{y}{x}, $$ to get the full $S^2$ metric in normal coordinates as $$ \frac{1}{\phi^2}\bigg(\Big( x^2+\frac{\sin^2\phi}{\phi^2}y^2\Big)dx^2 + 2\Big(1-\frac{\sin^2\phi}{\phi^2}\Big)xydxdy + \Big(y^2 + \frac{\sin^2\phi}{\phi^2} x^2\Big)dy^2\bigg). $$ This also recovers the result that $$ g_{11}(0, y) = \frac{\sin^2 y}{y^2}. $$