Well, it seems the two are the same. First, I assume we are not working in natural units, because of your final equation. Thus, the metric changes to $$ds^2=(cdt)^2-a^2\left [\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\right ]$$
Thus, your equation changes to $$c\frac{d^2t}{ds^2}+\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2$$
Second, note that, according to the resource, the velocity in the term $\frac{\dot{a}}{a}|u|^2$ is the three velocity. So, I'm going to denote the four velocity with $u^\mu$ and the three velocity with $u^i$. So, the equation given becomes $$\frac{du^0}{ds}+\frac{\dot{a}}{a}u^iu_i=0$$
where I have taken the liberty of changing out the first term, as the derivation is given within said resource, and you have already noted it.
We multiply by negative one, because the resource uses a different metric signature:
$$-\frac{du^0}{ds}-\frac{\dot{a}}{a}u^iu_i=0$$
Your first term is equal to $$-\frac{du^0}{ds}=-\frac{c}{c}\frac{du^0}{ds}=c\frac{d}{ds}\left(-\frac{1}{c}\frac{dt}{d\tau}\right)=c\frac{d}{ds}\left(\frac{dt}{ds}\right)=c\frac{d^2t}{ds^2}$$
For the second, we have $$\begin{align}-\frac{\dot{a}}{a}(u^i u_i)&=\frac{\dot{a}}{a}\left(\frac{a^2}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a^2r^2\left(\frac{d\theta}{ds}\right)^2+a^2r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2\right) \\
&=\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2
\end{align}$$
Therefore, $$\frac{du^0}{ds}+\frac{\dot{a}}{a}u^iu_i=0=c\frac{d^2t}{ds^2}+\frac{a\dot{a}}{1-kr^2}\left(\frac{dr}{ds}\right)^2+a\dot{a}r^2\left(\frac{d\theta}{ds}\right)^2+a\dot{a}r^2\sin^2\theta\left(\frac{d\phi}{ds}\right)^2=0$$
and they are the same.
$$ds^2 = y^{-2}dx^2 + y^{-2}dy^2.$$
I introduce a non coordinate orthonormal basis with one-forms
$$ds^2 = (\omega^{\hat{x}})^2 + (\omega^{\hat{y}})^2.$$
The new basis one forms can be identified as
$$\omega^{\hat{x}} = y^{-1}dx $$
$$\omega^{\hat{y}} = y^{-1}dy. $$
I calculate the exterior derivatives
$$d (\omega^{\hat{x}}) = d(y^{-1}dx) = - y^{-2} dy \wedge dx = - \omega^{\hat{y}} \wedge \omega^{\hat{x}} = \omega^{\hat{x}} \wedge \omega^{\hat{y}}$$
$$d (\omega^{\hat{y}}) = d(y^{-1}dy) = 0$$
$$d (\omega^{\hat{x}}) = -\Gamma^{\hat{x}}_{\;\hat{i}} \wedge \omega^{\hat{i}} = -\Gamma^{\hat{x}}_{\;\hat{x}} \wedge \omega^{\hat{x}} - \Gamma^{\hat{x}}_{\;\hat{y}} \wedge \omega^{\hat{y}} $$
$$\Gamma^{\hat{x}}_{\;\hat{y} } = - \omega^{\hat{x}}, \,\,\,
\Gamma^{\hat{x}}_{\;\hat{x} } = \omega^{\hat{y}}$$
$$\Gamma^{\hat{x}}_{\;\hat{y} \hat{x}} = -1, \,\,\, \Gamma^{\hat{x}}_{\;\hat{x} \hat{y}} = -1$$
$$\Gamma^{\hat{y}}_{\;\hat{y} } = - \omega^{\hat{y}}, \,\,\,
\Gamma^{\hat{y}}_{\;\hat{x} } = \omega^{\hat{x}}$$
$$\Gamma^{\hat{y}}_{\;\hat{y} \hat{y}} = -1, \,\,\, \Gamma^{\hat{y}}_{\;\hat{x} \hat{x}} = 1$$
Transformation matrices are:
$$
\Lambda^{\hat{a}}_{\;b} = \begin{bmatrix}
y^{-1} & 0 \\
0 & y^{-1} \\
\end{bmatrix}
$$
$$
(\Lambda^{-1}) ^{a}_{\;\hat{b}} = \begin{bmatrix}
y & 0 \\
0 & y \\
\end{bmatrix},
$$
hence
$$\Gamma^{x}_{\;yx} = (\Lambda^{-1}) ^{x}_{\;\hat{x}} \Gamma^{\hat{x}}_{\;\hat{y} \hat{x}}
(\Lambda) ^{y}_{\;\hat{y}} (\Lambda) ^{x}_{\;\hat{x}} = -y^{-1} $$
The other Christoffel symbols can be calculated in a similar way.
The solutions are correct.
$$\Gamma^{x}_{xx} = 0, \ \ \ \Gamma^{x}_{yx} = \frac{-1}{y}, \ \ \ \Gamma^{y}_{xx} = \frac{1}{y}, \ \ \ \Gamma^{y}_{yy} = \frac{-1}{y}$$
Best Answer
First of all, let's focus on the notation used in the question. The Schwarzschild metric is, in $(−+++)$ sign convention and units of $c=G=1$: $$ d{s}^2= -\left(1-\dfrac{2M}{r} \right) d{t}^2 + \left(1-\dfrac{2M}{r} \right)^{-1} d{r}^2 +r^2(d{\theta}^2+\sin^2{\theta}d{\phi}^2) $$ The reasons why it is convenient to put both the speed of light and Newton gravitational constant equal to $1$ are various and you can find an exhaustive list of the possible choices of the natural units in the following site.
Moreover, the convention of choosing the signature $(-+++)$ instead of the alternative $(+---)$ relies on the fact that it's the most used in modern textbooks on General Relativity and personally I find myself more comfortable with it. I'm sure you will be able to adjust this notation to your specific needs.
In our case, the covariant metric components form a diagonal matrix, namely: $$ g_{\mu\nu}=\begin{pmatrix} -\left(1-\dfrac{2M}{r} \right) & 0 & 0 & 0\\ 0 & \left(1-\dfrac{2M}{r} \right)^{-1}& 0 & 0\\ 0 & 0 & r^2 & 0\\ 0 & 0 & 0 & r^2\sin^2{\theta} \end{pmatrix} $$
In general, Christoffel symbols are related to the metric in the following way: $$ \Gamma^\sigma_{\mu \nu}=\dfrac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu} \right)\,\text{,} \qquad \qquad \partial_\mu \equiv \dfrac{\partial}{\partial x^\mu} $$
So, as I suggested in the comments below your question, instead of blindly calculating all $64$ symbols, we can use their symmetry properties and cleverly noting that some of them are identically equal to zero. For example, the Schwarzschild metric doesn't depend on $t$ or $\phi$.
Thus, by setting appropriately the index $\sigma =t$ from the equation above, also $\rho =t$ must be valid because the metric $g^{\sigma \rho}$ doesn't have off-diagonal elements. For the same reason, the last term $\partial_\rho g_{\mu\nu}$ is always zero for every value of $\mu$ and $\nu$ since nothing depends explicitly on time $t$. You have to then force $\mu=t$ in order to have diagonal elements of the metric and since this term only depends on $r$, there's only one possible choice for $\nu=r$. Doing this, you can compute $\Gamma^{t}_{rt}=\Gamma^{t}_{tr}$ and understand that all the other symbols for $\sigma=t$ are vanishing.
This might seem tricky and overcomplicated, but it allows us to discard by simple inspection all the symbols which are trivially equal to zero. An equivalent reasoning might be also implemented for $\sigma =\phi$ with the difference that now the element $g_{\phi\phi}$ depends both on $r$ and $\theta$, so we have two different non-zero symbols to consider.
After that, with a little work it's possible to identify the nonzero Riemann-Christoffel symbols: \begin{equation} \begin{matrix} \Gamma^r_{tt}=\dfrac{M(r-2M)}{r^3}\,\text{,} & ~ & \Gamma^r_{rr}=-\dfrac{M}{r(r-2M)}\,\text{,} & ~ & \Gamma^r_{\theta \theta}=-(r-2M)\,\text{,}\\ ~ & ~ & ~ & ~ & ~\\ \Gamma^r_{\phi \phi}=-(r-2M)\sin^2{\theta}\,\text{,} & ~ & \Gamma^t_{rt} = \dfrac{M}{r(r-2M)}\,\text{,} & ~ & \Gamma^\theta_{r\theta}=\dfrac{1}{r}\,\text{,}\\ ~ & ~ & ~ & ~ & ~\\ \Gamma^\theta_{\phi\phi}=-\sin{\theta}\cos{\theta}\,\text{,} & ~ & \Gamma^\phi_{r\phi}=\dfrac{1}{r}\,\text{,} & ~ & \Gamma^\phi_{\theta\phi} = \dfrac{\cos{\theta}}{\sin{\theta}}\,\text{.}\\ ~ & ~ & \end{matrix} \end{equation}