This question was posted a long time ago, but I will write my solution because I spent a lot of time stuck on this question.
$$\sigma(u,v) = ((r\cos(u)+a)\cos(v),(r \cos(u) + a )\sin(v), r \sin(u) ) $$
With $0<r<a$.
If $\gamma(t)= \sigma(u(t),v(t))$ is a geodesic parametrized by arc length, then $(u(t),v(t))$ must satisfy the differential equation (geodesic diffential equation)
$$v'' + 2\frac{(r\cos(u)+a)(r\cos(u)+a)' }{(r\cos(u)+a)^2} u' v' = 0 $$
$$u'' - \frac{(r\cos(u)+a) (r\cos(u)+a)'}{((r\cos(u)+a)')^2 + (\cos(u))^2} (v')^2 + \frac{r^2 \cos(u)\sin(u)}{((r\cos(u)+a)')^2 + (\cos(u))^2} (u')^2 = 0. $$
From the first equation, we conclude that
\begin{align*}
\frac{d}{dt}(r\cos(u(t)) +a)^2v'(t)) &= (r\cos(u)+a)^2 v''(t) + 2(r\cos(u(t)
+a)(r\cos(u)+a)' u' v'\\
&=(r\cos(u) + a)^2\left[ v'' + 2\frac{(r\cos(u)+a)(r\cos(u)+a)'}{(r\cos(u)+a)^2}u' v'\right] \\
&=0,
\end{align*}
which imply that $\exists$ $c \in \mathbb{R}$, such that
$$(r\cos(u(t) +a)^2 v'(t) = c.$$
Follows from these observations
\begin{align*}\cos(\theta(t)) &:= \frac{\langle \sigma_v(u(t),v(t)) , \gamma'(t)\rangle}{|\sigma_v(u(t),v(t))||\gamma'(t)|} \\
&= \frac{\langle \sigma_v , u'\sigma_u + v' \sigma_v \rangle}{\Vert\sigma_v\Vert}\\
&= v' \Vert \sigma_v \Vert \\
&= v'(t)\cdot \Vert(-(r\cos(u(t))+a)\sin(v),(r \cos(u(t)) + a) \cos(v(t)) , 0) \Vert\\
&= (r \cos(u(t)) +a)v'(t) = \frac{c}{r\cos(u(t)) + a }
\end{align*}
$$\Rightarrow (r \cos(u(t)) +a) \cos(\theta(t)) \equiv c. \qquad (1) $$
a)$\gamma$ is tangent to the parallel $u=\pi/2$
Suppose that $\exists$ $v_0,t_0 \in \mathbb{R}$ satisfying
\begin{align*}\gamma(t_0) &= \sigma\left(\left(\frac{\pi}{2}, v_0\right)\right) \\
\gamma'(t_0) &= \pm \frac{\sigma_v \left(\frac{\pi}{2},v_0\right)}{\left|\sigma_v\left(\frac{\pi}{2},v_0\right)\right|},\mbox{ remember that }\gamma \mbox{ is parametrized by arc length}.
\end{align*}
From (1)
$$c = (r \cos(u(t)) +a) \cos(\theta(t)) = (r \cos(u(t_0)) +a) \cos(\theta(t_0)) = a \quad (2) . $$
Supose by reductio ad absurdum that there is $t_1 \in \mathbb{R}$, satisfying $\gamma(t_1) = \sigma(u_1,v_1)$ with $u_1 \in [-\pi,\pi]\setminus [-\pi/2,\pi/2]$.
Using (2)
$$(r \cos(u(t_1)) +a) \cos(\theta(t_1)) = a, \hspace{0.1cm} \mbox{note that $0\leq \cos(\theta(t))\leq 1$} $$
$$\Rightarrow \cos(\theta(t_1)) = \frac{a}{(r \cos(u(t_1)) +a)}, $$
which implies $\cos(\theta(t_1)) > 1$. Contradiction!!
These arguments prove the first part of your question .
b) $\gamma$ intersect the parallel $u=0$ with angle $0<\theta<\pi/2$ .
$\exists$ $t_0$ $\in$ $\mathbb{R}$, such that
$$\gamma(t_0) = \sigma(0,v_0),\quad \cos(\theta(t_0))<\frac{a-r}{a+r} $$
From (1)
$$c = (r \cos(u(t)) +a) \cos(\theta(t)) = (r \cos(0) +a) \cos(\theta(t_0)) < a-r\qquad (3) $$
Suppose by reductio ad absurdum that $\gamma$ never reaches the parallel $u = \pi$, this is equivalent to say that $-\pi < u(t) < \pi$, $\forall$ $t$.
Note that $v'(t) \neq 0$, for all $t$. If $\exists$ $t_1$, such $v'(t_1) = 0$ $\Rightarrow$ $\cos(\theta(t_1)) = 0$ $\Rightarrow$ $\gamma'(t_1)$ is l.d. with a meridian, by the unicity of the geodesics (given a point and a direction there is a unique geodesic passing through this point) $\gamma$ must be the meridian, it implies that $\cos(\theta(t_0)) = 0$ $\Rightarrow$ $\theta(t_0) = \theta = \pi/2$. Contradiction! So $v'(t) \neq 0$, $\forall$ $t$.
Suppose without loss of generality that $v'(t) >0$ $\forall t$.
On the order hand $u'(t) \neq 0$. Suppose that $\exists$ $t_1$ such that $u'(t_1) = 0$, it implies that $\cos(\theta(t_1)) = 1$ (remember that we suppose that $-\pi < u(t)< \pi)$. But
$$c = (r\cos(u(t)) + a )\cos(\theta(t)) = (r \cos(u(t_1))+a) 1 > a-r $$
contradicts (3). So $u'(t) \neq 0$ $\forall$ $t$.
Suppose without loss of generality that $u'(t) >0$ $\forall t$.
Note that by the hypothesis $\gamma$ is parametrized by arc length
$$ |u'| |\sigma_u| + |v'| |\sigma_v| = 1 $$
Using that
$$|\sigma_u| = r, \quad |\sigma_v| > a-r>0 $$
this implies that
$$0< u'(t) \leq 1/r , \qquad 0< v'(t) < 1/(r-a) \quad (4)$$
Now we can conclude that the domain of $\gamma$ contains $[t_0, \infty)$
. If this not occur exists a maximal $\omega^{+}$ $\in$ $\mathbb{R}$ such that $\gamma: [0, \omega^+) \rightarrow \mathbb{R}^3$. Remember that $(u(t),v(t))$ resolves the geodesic differential equation, by ODE theorems we know that $(u(t),v(t))$ $\to$ $\partial \mathbb{R}^2$ when $t \rightarrow \omega^{+}$ (This means that for every compact $K$ $\subset$ $\mathbb{R}^2$, $\exists$ $t_K$ such that $(u(t),v(t)) \not\in K$, forall $t>t_K$).
so $$|(u(t),v(t))| \rightarrow \infty, \quad \mbox{when}\quad t \rightarrow \omega^{+}. $$
But from Mean Value Theorem and (4).
$$|u(t) - u(t_0)| < \frac{\omega^{+} - t_0}{r} \quad |v(t) - v(t_0)| < \frac{\omega^{+}- t_0}{a-r}, \quad \forall t \in [t_0, \omega^{+})$$
Contradiction! Then $\gamma$ is well defined for all $t$ $\in$ $[t_0, \infty).$
Note that $u'(t) >0$, and (by the reductio ad absurdum hypothesis) $u(t) < \pi$, $\forall$ $t$ $\in$ $[t_0, \infty]$ as $u (t)$ is limited and increasing, there is $\tilde{u} \leq \pi$ satisfying
$$\lim_{t \rightarrow \infty} u(t) = \tilde{u} $$
Now we will prove that $\liminf\limits_{t \rightarrow \infty} u'(t) \rightarrow 0 $ when $t \rightarrow \infty$. In case otherwise, there will be $ c> 0 $ such that $ u '(t)> c $ for all $ t> t_c $ which implies that
$$ | u (t) - u (t_c) | > c | t - t_c | $$
$$\Rightarrow u(t) \rightarrow \infty, \qquad \mbox{when} \quad t \rightarrow \infty $$
Contradiction! So $\exists$ sequence $(t_k)_{k \in \mathbb{N}}$, such that $t_k \rightarrow \infty$ and $u'(t_k) \rightarrow 0$ when $k\rightarrow \infty$. But this imply that
$$\cos(\theta (t_k)) \rightarrow 1, \quad \mbox{when}\quad k \rightarrow \infty $$
and
$$a-r > c = \lim_{k \rightarrow \infty} (r\cos(u(t_k)) +a) \cos(\theta (t_k)) = (r \cos(\tilde{u}) +a ) 1 \geq a-r $$
(Final) Contradiction! So we conclude that $u(t)$ must pass to $\pi$ (or $-\pi$, if you suppose that $u'(t) <0$).
Therefore, we proved the theorem.
