Setup: $S^2 \subset \mathbb{R}^{3}$ is the unit sphere and for fixed $z \in (0,1)$ define $c(t)=(\sqrt{1-z^2}\cos t, \sqrt{1-z^2}\sin t, z)$.
Let $V(t)$ be the vector field on $S^2$ (along $c$) given by $V(t)=c'(t)=(-\sqrt{1-z^2}\sin t, \sqrt{1-z^2}\cos t,0)$.
We want to compute the covariant derivative $\nabla_{c'(t)}c'(t)=\frac{DV}{dt}$ defined by projection of $\frac{dV}{dt}$ onto $T_{c(t)}S^2$
I have computed that the tangent space $T_{c(t)}S^2$ is the space of vectors perpendicular to $c(t)$ i.e. $(c(t))^{\perp}$.
I am now struggling with actually projecting the vectors $\frac{dV}{dt}=(-\sqrt{1-z^2}\cos t, -\sqrt{1-z^2}\sin t,0)$ onto the plane $T_{c(t)}S^2$. Originally I thought I can project $\frac{dV}{dt}$ onto $c'(t)$, which is a vector that lies in $T_{c(t)}S^2$ (in which case I get $0$) but now I'm not sure I can do this.
Let $c: I \to \Bbb S^2$ be a smooth curve and $N\colon \Bbb S^2 \to \Bbb R^3$, $N(p) = p$ the unit outward normal field. By definition, we have $$\begin{align}\frac{{\sf D}V}{{\rm d}t}(t) &= {\rm proj}_{T_{c(t)}\Bbb S^2} V'(t) \\ &= V'(t) - \frac{\langle V'(t), c(t)\rangle}{\langle c(t),c(t)\rangle}c(t) \\ &\stackrel{(1)}{=} V'(t) - \langle V'(t),c(t)\rangle c(t) \\ &\stackrel{(2)}{=} V'(t) + \langle V(t),c'(t)\rangle c(t),\end{align}$$where in (1) we use that $c(t) \in \Bbb S^2$ means $\langle c(t),c(t)\rangle =1$, and in (2) we use that $\langle V(t), c'(t)\rangle = 0$ since $V(t) \in T_{c(t)}\Bbb S^2 = (c(t))^\perp$ (hence we can pass the derivative to the other argument of the inner product, swapping the sign). Note that for $V(t) = c'(t)$ we obtain $$\frac{{\sf D}c'}{{\rm d}t}(t) = c''(t) + \langle c'(t),c'(t)\rangle c(t).$$
In your case $$\begin{align} c(t) &= (\sqrt{1-z^2}\cos t, \sqrt{1-z^2} \sin t, z)\quad\mbox{and} \\ V(t) &= c'(t) = (-\sqrt{1-z^2}\sin t, \sqrt{1-z^2}\cos t, 0)\end{align}$$gives $$c''(t) = (-\sqrt{1-z^2}\cos t, -\sqrt{1-z^2}\sin t, 0)$$ and so$$\begin{align} \frac{{\sf D}c'}{{\rm d}t}(t) &=(-\sqrt{1-z^2}\cos t, -\sqrt{1-z^2}\sin t, 0) + (1-z^2)(\sqrt{1-z^2}\cos t, \sqrt{1-z^2} \sin t, z) \\ &=(-z^2 \sqrt{1-z^2}\cos t,-z^2 \sqrt{1-z^2}\sin t,z-z^3).\end{align}$$Simplify further as needed.