This answer relates to the first bullet point of the OP.
(The second bullet point has been answered in a note due to Arthur. )
The question can be interpreted in the case of the Euclidean and in the case of Hyperbolic geometry.
Euclid defines the parallel to a straight the following way: let $a$ and $P$ be a straight and a point, resp. in a plane. The straight $b$ through $P$ is parallel to $a$ if it does not have a common point with ot. (An equivalent formulation of Euclid's parallel postulate claims hat there is only one parallel through $P$ to $a$.
Bolyai and Lobachevsky define the parallel to $a$ through $P$ the following way: Construct a straight through $P$ in the common plane of $a$ and $P$. Let $b$ intersect $a$ then turn $b$ around $P$ so that the intersection point gets further and further away from $P$, there will be a situation when $b$ does not meet $a$ the first time. Then there will be a situation when $b$ intersect $a$ again the first time. (This latter definition allows that $a$ has two parallels through $P$. The hyperbolic parallel postulate claims that there are two parallels to through $P$ to $a$.)
In both cases the essence of parallelism is that parallel lines do not meet.
Now, consider the Klein model of hyperbolic geometry. This model embeds the hyperbolic plane (space) in the Euclidean plane (space) by taking am open disk (sphere) whose (internal) points are the point of the hyperbolic plan (space) and whose chords are the straights.
The following figure depicts the Klein model of the hyperbolic plane.
Here $P$ is a hyperbolic point, $a$, $b$, and $b'$ are parallel hyperbolic straight lines to $a$ in the Bolyay-Lobachevsky sense. Apparently the two parallels meet on the circle (in the infinity). There is no contradiction because the circle itself does not belong to the model of the hyperbolic plane. But it is so good to be mystic and say that the parallels meet in the infinity.
But what about the parallels of the Euclidean geometry that do not meet but in the infinity?
It is harder to intuit the Euclidean parallels as pairs of straight lines meeting in infinity. It helps if one can imagine the Euclidean plane modeled within itself the following way.
Take the inside of an Euclidean circle and remove its center. Now, consider the Euclidean circles that go through the center and are within the circle. These Euclidean circles will model the Euclidean straight lines. Every pair of straight lines meet in the center (not belonging to the model plane). Some of these model straight have another meeting point and some don't. In the figure above $a$ and $b$ are parallels having only one "common point in the infinity". Now, we know that the center of the big circle represents the point in the infinity (not belonging to the model), in which point parallels meet.
Note. Actually the inverse images of the Euclidean straights wrt the black circle are the images of the Euclidean straight.
Best Answer
This is not true in hyperbolic geometry. It is easy to see a counterexample in Poincare half-plane model (https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model). There, the points are the points of the upper half-plane $\{(x,y)\in\mathbb R^2\mid y>0\}$, and we have two types of lines: straight lines orthogonal to $x$-axis, so sets given by $\{(x_0,y)\mid y>0\}$ for any $x_0\in\mathbb R$ are lines, and upper parts of the circles with centers on $x$-axis, i.e. sets given by $\{(x,y)\mid (x-x_0)^2+y^2=r^2,y>0\}$ for $x_0\in\mathbb R$ and $r>0$ are lines.
Consider lines $p$ and $q$ given by: $p=\{(x,y)\mid x^2+y^2=1,y>0\}$ and $q=\{(x,y)\mid (x-2)^2+y^2=1,y>0\}$; they don't intersect. Points between $p$ and $q$ are all points outside of these circles. Take e.g. $M=(0,3)$. It is easy to see that lines $QM$ don't intersect $p$ when $Q\in q$.
Also, you can easily see an counterexample in Beltrami-Klein model (https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model).