I'm really struggling with calculating an example of the levi-civita derivative right now. I'm assuming a curve on a surface $M \subseteq \mathbb{R}^3$, then the levi-civita derivative of a vectorfield $Y$ along $M$ wrt to a vectorfield $X$ on $M$ is given by $\nabla_X Y := (D_X Y)_{tangential}$ where $D_X Y$ is the directional derivative. This directional derivative of a vector field wrt to another one is defined as follows: if $Y = y^k \partial_k$ then $D_X Y = \sum_k (X(y^k)) \partial_k$ - so we apply $X$ as a differential operator to the component functions $y^k$. This seems to make sense but I'm struggling with applying it.
Consider the parameterized curve $$\gamma(t) = (t², t³),$$ then we should be able to compute the levi-civita derivative of its velocity vector field along itself. If I'm not mistaken this should end up just being the second derivative $\gamma''$ of $\gamma$ wrt to $t$.
Here's my attempt and thinking:
So our manifold $M$ is given by $\gamma(\mathbb{R})$. The velocity vector field $V_p$ is defined as follows: given some point $p = (x,y) = (t^2,t^3)$ in $M$ we define $V_p := (2t, 3t^2) = \gamma'(t)$. From this point on I'm a bit unsure: so we have $V_p = 2t \partial_x + 3t^3 \partial_y$ (?), our component functions are $v^x(x,y) = 2t = 2 \sqrt[3]{y}$ and $v^y(x,y) = 3t^2 = 3 \sqrt[3]{y}^2$. We now want to apply $2t \partial_x + 3t³ \partial_y$ to these functions. We find: $$V_p v^x(x,y) = 2t \cdot 0 + 3t^3 \cdot \frac{2}{3 \sqrt[3]{y}^2} = \frac{2 y}{\sqrt[3]{y}^2} = 2 \sqrt[3]{y}$$ and
$$V_p v^y(x,y) = 2t \cdot 0 + 3t^3 \cdot \frac{2}{\sqrt[3]{y}} = \frac{6 y}{\sqrt[3]{y}} = 6 \sqrt[3]{y}^2.$$
So the derivative $D_{V_p} V (x,y) = (2 \sqrt[3]{y}, 6 \sqrt[3]{y}^2) = 2 \sqrt[3]{y} \partial_x + 6 \sqrt[3]{y}^2 \partial_y$ or expressed in terms of $t$ we have $D_{V_p} V (t) = (2t, 6t^2)$? And from that point on I'd have to find the tangential component?
This just seems all wrong and I'm confused - I thought this levi-civita derivative should in this case just end up being the second derivative of the curve wrt to $t$, so $y''(t) = (2, 6t)$ but I somehow have one $t$ too much? And when doing to projection things get even worse.
In case it helps: I'm working with Tu's differential geometry and an introduction to smooth manifolds.