To make things easier, consider $[0,a]$. The function $x^2$ is increasing over $[0,a]$ and it is continuous there. Thus, if we partition $[0,a]$ with $$P=\{t_0,\dots,t_n\}$$ we will have $$M_i=t_i^2$$
and $$m_i=t_{i-1}^2$$
since a continuous functions attains it maximum and minimum in a closed interval. Thus
$$L(f,P)=\sum_{i=1}^n t_{i-1}^2(t_i-t_{i-1})$$
$$U(f,P)=\sum_{i=1}^n t_{i}^2(t_i-t_{i-1})$$
As you say, consdier the partition where $$t_i=\frac{ia}{n}$$
Then $$\eqalign{
& L(f,P) = \sum\limits_{i = 1}^n {{{\left( {i - 1} \right)}^2}\frac{{{a^2}}}{{{n^2}}}} \frac{a}{n} = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{{\left( {i - 1} \right)}^2}} \cr
& U(f,P) = \sum\limits_{i = 1}^n {{i^2}\frac{{{a^2}}}{{{n^2}}}} \frac{a}{n} = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} \cr} $$
Since $$\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$$ we get $$\eqalign{
& L(f,P) = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 0}^{n - 1} {{i^2}} = \frac{{{a^3}}}{3}\frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{{2{n^3}}} \cr
& U(f,P) = \frac{{{a^3}}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} = \frac{{{a^3}}}{3}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2{n^3}}} \cr} $$
from where $$\eqalign{
& L(f,P) \to \frac{{{a^3}}}{3} \cr
& U(f,P) \to \frac{{{a^3}}}{3} \cr} $$
Then $$\int\limits_a^b {{x^2}dx} = \int\limits_0^b {{x^2}dx} - \int\limits_0^a {{x^2}dx} = \frac{{{b^3}}}{3} - \frac{{{a^3}}}{3}$$
Observe that through the map
$$\left(\begin{array}{c}
u^1\\
u^2
\end{array}\right)
\stackrel{f}\to
\left(\begin{array}{c}
x^1\\
x^2\\
x^3
\end{array}\right),
$$
you have a GPS for points in the surface. It is with this that you get $\frac{\partial f}{\partial u^1}$ and $\frac{\partial f}{\partial u^2}$, which are the two columns of the jacobian $Jf$, and serve to define the tangent space at any point on the surface.
So to travel along a curve in the surface you need to travel along a curve in the domain $U$. This implies that $u^1,u^2$ depends on a parameter, which we can denote with $t$.
Schematically
$$t\stackrel{\gamma}\to
\left(\begin{array}{c}
u^1(t)\\
u^2(t)
\end{array}\right)
\stackrel{f}\to
\left(\begin{array}{c}
x^1(t)\\
x^2(t)\\
x^3(t)
\end{array}\right),
$$
will give the way to control the positions on curve over the surface.
Now $\dot{c}=\frac{d}{dt}(f\circ\gamma)=Jf\cdot\dot{\gamma}$ and since $\dot{\gamma}=\dot{u^1}e_1+\dot{u^2}e_2$ then
$$\dot{c}=Jf(\dot{u^1}e_1+\dot{u^2}e_2)=\dot{u^1}Jf(e_1)+\dot{u^2}Jf(e_2),$$
so
$$\dot{c}=\dot{u^1}\frac{\partial f}{\partial u^1}+\dot{u^2}\frac{\partial f}{\partial u^2}.$$
This way you can use it in $\nabla_{\dot{c}}\dot{c}$ to get
$$\nabla_{\dot{c}}\dot{c}=\sum_s(\ddot{u^s}+\sum_{ij}\Gamma^s{}_{ij}\dot{u^i}\dot{u^j})\frac{\partial f}{\partial u_s}.$$
This last formula is get as follow:
$$\nabla_{\dot{c}}\dot{c}=
\nabla_{\sum_s\dot{u^s}\frac{\partial f}{\partial u_s}}
\sum_{\sigma}\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\quad=\sum_{s{\sigma}}\dot{u^s}\nabla_{\frac{\partial f}{\partial u_s}}
\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{s{\sigma}}\dot{u^s}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}}\dot{u^s}\dot{u^{\sigma}}\nabla_{\frac{\partial f}{\partial u_s}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{s{\sigma}}\dot{u^s}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s{\sigma}}\frac{\partial f}{\partial u_r},$$
$$\qquad=\sum_{s{\sigma}}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{du^s}{dt}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s\sigma}\frac{\partial f}{\partial u_r},$$
$$\qquad=\sum_{{\sigma}}
\frac{d^2u^s}{dt^2}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s\sigma r}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{{\sigma}}\left(
\frac{d^2u^{\sigma}}{dt^2}+\sum_{sr}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\right)
\frac{\partial f}{\partial u_{\sigma}}.$$
Best Answer
Something is wrong in the lecture statement. Let $n=1,a_1=5,t_1=3.$ And replace $b$ by $x$ so it is a function of $x$ and usual derivatives are the partial. Then $$A=\frac{5}{(1+x)^3}.$$ Now compute $$-\frac{1}{A} A'(x) = \frac{3}{1+x}.$$ But this does not match the right side of the lecure statement, which is $$-\frac{t_1 a_1}{(1+x)^{-t_1}}=-\frac{15}{(1+x)^{-3}}.$$ It looks too different for the lecture statement to be right, even if in you question there was a simple sign error somewhere.
By the way, your first step looks right.