Vector fields pointing from one fiber to another - Notation Let $(P,M,\pi)$ be a Principal $G$ bundle.
I am quoting following lines from the book Modern Differential geometry for Physicists by C J Isham.

The basic idea of parallel transport/covariant differentiation is to compare points in neighbouring fibers ina way that is not dependent on any particular local bundle trivialisation. This suggests looking for vector fields that point from one fiber to another.

I do not understand why does idea of comparing fibers of two nearby points should suggest looking for some vector fields. It is not clear what does it mean to say vector field that points from one fiber to another.
I thought I might understand if I go further but still it is not clear.
In search of some vector fields (which is not clear at this point what kind of vector fields they are) he defines vector field $X^A$ on $P$ given an element $A\in T_eG$. I am not defining it because definition is not relavent here. He then says with out proof that $$\pi_*\left( X^A_p\right)=0$$
I have checked this and it turns out to be true. Before saying this, he says 

How ever, the particular vector fields are not suitable for our purposes since they do not point from one fiber to another. On the contrary, the vectors $X^A_p$ are tangent to the fiber at $p\in P$ and hence they point along the fiber, rather than away from it. Technically, $X^A_p$ is said to be a vertical vector i.e., it belongs to the vertical subspace $V_pP$ of $T_pP$ defined by 

$$V_pP= \left\{ \tau\in T_pP : \pi_*\tau=0\right\}$$ 
where $\pi$ is the projection map of the bundle.
I do not understand what does it mean to say vectors are tangent to the fibers or point along the fibers. 
 A: Locally a fiber bundle looks like a product. So to understand local statements about fiber bundles, it is enough to study a product. For example, a line bundle over a 1-manifold looks like the trivial bundle $\mathbb{R}^2\overset{\pi}{\to}\mathbb{R}$, a bundle whose total space is the product $\mathbb{R}\times\mathbb{R}$, whose base is $\mathbb{R},$ and projection $\text{proj}_2\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ which has value $\text{proj}_2(x,y)=x$. 
In this trivialization, a point in the base looks like $(3,0)$, a fiber looks like $\pi^{-1}(3)=3\times\mathbb{R}=\{(3,y)\colon y\in\mathbb{R}\}.$  A vertical vector looks like $e_y=\partial_y$, and this vector is tangent to the fiber $\pi^{-1}(3).$ A vector pointing to an adjacent fiber may look like $e_x=\partial_x$ or $e_x+e_y=\partial_x+\partial_y$. Anything which is not vertical. The subbundle of vertical vectors is the set of all vectors $V(\mathbb{R}^2)=\{f(x,y)\partial_y\}.$
A: It is indeed incorrect to say a vector field along a fiber is pointing to another point on another fiber. Just like a a tangent vector at a point on a sphere isn't pointing to another point on the sphere; it isn't because it's in the tangent space. That said, one needs to create a a means to transport vectors, otherwise calculus on manifolds couldn't be done. That's why connection 1-forms exist (also why Lie derivatives exist).
