Why can we always take the zero section of a vector bundle? $\require{AMScd}$
As I understand it, a rank $k$ vector bundle is a pair of topological spaces with a map between them
$$
E\xrightarrow{p}B
$$
such that there exists an open cover $(U_\alpha)$ of $B$ over which $E$ locally looks like $U_\alpha\times\mathbb R^k$.  Precisely, we have homeomorphisms
$$
\phi_\alpha\colon p^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k
$$
such that the diagram
\begin{CD}
p^{-1}(U_\alpha) @>\phi_\alpha>> U_\alpha\times\mathbb R^k\\
@VVbV @VV\text{pr}_1V\\
U_\alpha @>\text{id}>> U_\alpha
\end{CD}
commutes (I can't do diagonal arrows).  
A section of the vector bundle is a continuous map $s\colon B\to E$ such that $p\circ s=\text{id}_B$.  A basic fact about vector bundles is that every vector bundle admits at least one section, namely the zero section, obtained by identifying each $p^{-1}(U_\alpha)$ with $U_\alpha\times\mathbb R^k$ and taking $s(b)=0$.  Explicitly, we are taking:
$$
s(b)=\phi_\alpha^{-1}((b,0))
$$
for $b\in U_\alpha$.  
But in order to get a well defined continuous map, we need to ensure that $\phi_\alpha^{-1}(b,0)=\phi_\beta^{-1}(b,0)$ if $b\in U_\alpha\cap U_\beta$.  Now one thing we certainly can't do is require that $\phi_\alpha$ and $\phi_\beta$ actually agree on the intersection - then we could glue them all together to get a global trivialization of the bundle, which we can't do in general.  Why then are we able to insist that they agree at a certain point?  To me, it seems equivalent to the existence of a global section, and hence completely circular.  

There must be something obvious I'm missing.  In the meantime, here's an equivalent formulation of the problem: we define an affine bundle $E\xrightarrow{p}B$ to be the same as a vector bundle, but we identify the fibres with $k$-dimensional affine linear subspaces of some fixed real vector spaces.  So they needn't contain a special zero point.  
Since affine spaces are homeomorphic to vector spaces of the same dimension, any rank $k$ affine bundle is isomorphic to some rank $k$ vector bundle (though I'm not at all sure about whether this is true, for similar reasons).  So an affine bundle should admit a global section, but there is no easy way to see how to get one.  
 A: I think it's more straightforward than you're currently thinking.  Every fiber $F_b$ has the structure of a vector space, and therefore has a zero element $0_b$.  Sending an element $b$ to the zero vector $0_b$ defines the (continuous) zero section you're after.
A: As was explained in the other answers, for vector bundles one requires that the fibers are equipped with a vector space structure, and this immediately gives a (continuous) section. For affine bundles (in your sense), as you point out, there is no preferred origin so one cannot just define the zero section using the local charts. Nonetheless, assuming the existence of partitions of the unity for the base one can construct a section as follows.
Fix a (locally finite) affine bundle atlas $\{\phi_i:U_i\times A\rightarrow E\}$ and let $\{\rho_i\}$ be a partition of the unity associated to the covering $\{U_i\}$ of $B$. In an affine space convex combinations of vectors do not depend of the base point, so you simply define
$
s(x)=\sum \rho_i(x)\phi_i(x,0).
$
In this way you get a continuous section. 
A: Viktor Kleen is right that your definition is missing something, but I think the usual formulation of what you're missing is a bit more than what Viktor wrote.  Part of the structure of a vector bundle is an assignment, to each fiber, of a vector-space structure on that fiber; furthermore the local coordinate maps $\phi_\alpha$ in your definition are required to be linear on every fiber.  
The vector-space structure on the fibers includes, in particular, a choice of zero vector in every fiber, so you have a function assigning to each point $b\in B$ the zero vector of the fiber over $b$.  This function is continuous on each $U_\alpha$ because it corresponds, under the homeomorphism $\phi_\alpha$, to the continuous function $U_\alpha\to U_\alpha\times \mathbb R^k$ sending each $b$ to $(b,0)$ (where $0$ means the zero vector in $\mathbb R^k$).  Finally, a function continuous on each set $U_\alpha$ of an open cover is continuous on the whole space, so you have a continuous zero section.
A: In your definition you are missing the condition that the transition maps for a vector bundle are supposed to be linear fiberwise. Once you assume that, the problem you're facing in gluing will go away.
A: If you had a canonical zero element in each fiber, you would see the section right? So but that is already true, since you usually require in the definition that your fibers admit a vector space structure and you also require that the local trivializations are isomorphisms i.e. also preserve zero. That gives you a continous choice of zeros everywhere.
