From what I can tell, the error arises because you need to reparametrize your $u_{1}$ curve to have constant speed. Once you do this, the geodesic equations should work out.
Denote the metric by $Edu_{1}du_{1} + Gdu_{2}du_{2}$ (i.e. $E = g_{11}$ and $G = g_{22}$ with your notation above) and observe that $E$ and $G$ depend on only $u_{1}$. As you have expressed above, the geodesic equations reduce to
\begin{align*}
u_{1}^{\prime\prime} + \Gamma^{1}_{11}\left(u^{\prime}\right)^{2} + \Gamma^{1}_{22}\left(v^{\prime}\right)^{2} &= 0\\
u_{2}^{\prime\prime} + \Gamma^{2}_{1 2}u_{1}^{\prime}u_{2}^{\prime} &=0\\
\end{align*}
Since the class of curves under consideration is the meridians, you have correctly observed that all terms involving a factor of $u_{2}^{\prime}$ will vanish. The remaining geodesic equation(s) is
\begin{equation}
u_{1}^{\prime\prime} + \Gamma^{1}_{11} \left(u^{\prime}\right)^{2} = 0,
\end{equation}
where $ \Gamma^{1}_{11} = \frac{f^{\prime}f^{\prime \prime}}{1 + (f^\prime)^2} = \frac{E_{u_{1}}}{{2E}}$
Now given a curve curve of the form $\gamma(t) = \sigma\left(t, c\right)$ (i.e. $u_{1}(t) = t$), we should first reparametrize $\gamma$ to have constant speed. Observe that the arc length function is
$$s(t) = \int\limits_{0}^{t} \vert \vert \gamma^{\prime}(w) \vert \vert dw = \int \limits_{0}^{t} \sqrt{E(w)} dw,$$
where the last equality follows from the fact that the tangent vector of our meridian curve is tangent to the $u_{1}$ curves. Of course, the fundamental theorem of calculus implies that $\frac{ds}{dt} = \sqrt{E(t)} > 0$, and your meridian admits a unit speed reparametrization via the inverse function $t = t(s)$.
Take the unit speed reparametrization to be $\alpha(s) = \sigma(t(s), c)$, (i.e. $u_{1}(s) = u_{1}(t(s))$ where $u_{1}(s) = t(s)$ and note that by the chain rule (and the fact that you have a meridian so you are tangent to a $u_{1}$ curve) you obtain the following derivatives:
$$\frac{du_{1}}{ds} = \frac{du_{1}}{dt}\frac{dt}{ds} = 1\cdot\frac{1}{\sqrt{E}},$$
and
$$\frac{d^{2}u_{1}}{ds^2}= -\frac{E_{u_{1}}}{2E^{2}}.$$
You should now be able to observe that the the unit speed parametrization of your meridian satisfies the geodesic equation above.
Finally, a quick remark regarding the importance of the parametrization. Take the Euclidean plane with metric $ds^{2} = dx^{2} + dy^{2}$ (i.e. with $(g_{ij}) = \begin{pmatrix}
1&0\\
0&1\\
\end{pmatrix}$). If you write down the geodesic equations with the Christoffel symbols, you end up with
$$ x^{\prime \prime} = 0 \hspace{.5in} \textrm{and} \hspace{.5in} y^{\prime \prime} = 0.$$
Now consider the line parametrized by $\gamma(t) = (t^{3}, t^{3})$. The curve $\gamma$ clearly traces out a geodesic, but the given parametrization does not satisfy the geodesic equations.
The world lines that extremize the proper time between A and B are those which satisfy the Lagrange equation:
\begin{equation}
-\frac{d}{d \sigma} \left(\frac{\partial L}{\partial({dx^{\alpha} / d \sigma})} \right)+\frac{\partial L}{\partial x^{\alpha}}=0
\end{equation}
Where the Lagrangian is:
\begin{equation}
L= \left(-g_{\alpha \beta} \frac{d x^{\alpha}}{d \sigma} \frac{d x^{\beta}}{d \sigma} \right)^{\frac{1}{2}}
\end{equation}
The Schwarzschild metric in geometrical units is
\begin{equation}
\left(
\begin{array}{cccc}
-(1-\frac{2 m}{r}) & 0 & 0 & 0 \\
0 & (1-\frac{2 m}{r})^{-1} & 0 & 0 \\
0 & 0 & r^2 & 0 \\
0 & 0 & 0 & r^2 \sin ^2(\theta ) \\
\end{array}
\right)
\end{equation}
The parameter t is therefore called world time. If m = 0 the Schwarzschild metric reduces to the Minkowski metric; if $m \neq 0$ it is singular for $r = 0$. For r large with respect to m the Schwarzschild metric coincides with the Newtonian approximation, since $2m/r$ is the Newtonian potential outside a spherical body of mass m centered at $r = 0$. In addition, when
$1−(2m/r)$ vanishes for $r = 2m$, the metric appears to be singular there. This singularity can be removed by choosing a different coordinate system.
The geodesic equations based on the Schwarzschild metric above are
\begin{equation}
\left(
\begin{array}{ccc}
\frac{d u_0}{d \tau} = \frac{2 m u_0 u_1}{2 m r-r^2} \\
\frac{d u_1}{d \tau} = \frac{m (2 m-r) u_0^2}{r^3}-\frac{m u_1^2}{2 m r-r^2}-(2 m-r) u_2^2-(2 m-r) \sin ^2(\theta) u_3^2 \\
\frac{d u_2}{d \tau} = \cos (\theta) \sin (\theta ) u_3^2-\frac{2 u_1 u_2}{r} \\
\frac{d u_3}{d \tau} = -\frac{2 (u_1+r \cot (\theta ) u_2) u_3}{r} \\
\end{array}
\right)
\end{equation}
where $u_{\alpha}$ are the component of the four proper velocity vector while $\tau$ is the proper time. When looking at the metric you can see that is independent of $t$, thus there is a Killing vector associated with this symmetry under displacement of the time coordinate t with components $\psi=(1,0,0,0)$. The first equation can be written as:
\begin{equation}
\frac{d}{d \tau} \left((1-\frac{2 m}{r}) u_0(\tau) \right) = 0
\end{equation}
which is integrated to give
\begin{equation}
\left((1-\frac{2 m}{r}) u_0(\tau) \right) =- g_{00} \frac{dt}{d \tau}= E
\end{equation}
with E constant.
Now you have to remember that the dot product is obtained by computing the first fundamental form.The first fundamental form is the restriction of the usual dot product in $\mathbb{R}^3$ to the tangent plane $T_p S$. Given two vector a and b
the first fundamental form $Ip (a, b)$ at the point $p$ can be expressed as the bilinear form $a^T g b$, where g is the matrix. In the case of the xy-plane or a cylinder $g$ is a diagonal matrix with diag(g) = (1,1). In this case the dot product of $a$ and $b$ is
\begin{equation}
(a_1,a_2) \left(
\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}
\right) (b_1,b_2)^T= a_1 b_1+ a_2 b_2
\end{equation}
However is the matric is not flat we have to account for the terms $g_{\alpha, \beta}$. So yes, $U^aV_a$ is a dot product, with the index a running from 1 to the 4, since you are in a 4 dimensional space. This dot product can be written as
\begin{equation}
g_{00} u_0 v_0-g_{00}^{-1} u_1 v_1+r^2 u_2 v_2+ r^2 \sin^2(\theta)u_3 v_3
\end{equation}
However, The 4-velocity of a stationary observer is $(u_0,0,0,0)$, so for a stationary observer we have only the term $g_{00} u_0 v_0$.
Consider now the lapse of proper time τ between two events at a fixed spatial point (differentiation with respect the position is zero) in Schwarzschild space-time:
\begin{equation}
ds^2=-d \tau^2=-g_{00} dt^2
\end{equation}
hence
\begin{equation}
d \tau=\sqrt{g_{00}} dt
\end{equation}
The element of proper time dτ is measured by a clock at the particular point, while the element of world time dt is fixed for the whole manifold. Since $g_{00}<1$, so $d \tau < dt$ i.e. clocks go slower in a gravitational field. In fact time itself goes slower in a gravitational field. In flat spacetime $m$ turn down to zero so that $dt=d \tau$, and a clock records the coordinate time $t$.
Best Answer
A short answner:
For your first question, here a hint: consider that $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ where $\beta = v/c$, approximate $\gamma$ when $\ v << c$ (Slow moving particle).
for the second and third question's:
yes, $\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}=\gamma^2 u^i u^j$, $\gamma^2 \approx 1$, then $\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $\Gamma_{00}^a\frac{dx^0}{d\tau}\frac{dx^0}{d\tau}$, but $\frac{dx^0}{dt}=1$, then $\frac{d^2x^a}{d\tau^2}+\Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ \gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.