What does the stalk of a sheaf of discontinuous sections look like? I'm having some kind of cognitive dissonance here, but I'm having trouble figuring out which of my beliefs is false.  Let $\mathscr{F}$ be a sheaf of abelian groups on a topological space $X$ and $x \in X$.  Write $\mathscr{G} := \prod_{x \in X} i_\ast(\mathscr{F}_x)$, where $i_\ast(\mathscr{F}_x)$ is the $\mathscr{F}_x$-valued skyscraper sheaf at $x$.
We have $\mathscr{G}(U) := \prod_{x \in X} \Gamma(U, i_\ast(\mathscr{F}_x)) = \prod_{x \in U} \mathscr{F}_x$, which we can regard as the set of functions $s : U \to \bigsqcup_{x \in U} \mathscr{F}_x$ by writing $s(x)$ for the $x$-entry of the direct product.  So $\mathscr{G}$ is the sheaf of discontinuous sections of $\mathscr{F}$.
Since taking stalks commutes with taking direct sums products, we have $\mathscr{G}_x = \mathscr{F}_x$ for each $x \in X$.
On the other hand, looking at elements of $\mathscr{G}_x$ directly, they represent functions $s : U \to \bigsqcup_{x \in U} \mathscr{F}_x$ up to equivalence on some open set.  But it's easy to construct all sorts of such functions which don't agree on any open neighborhood of $x$ but which have the same value for $s(x)$ -- just send all the neighboring points absolutely anywhere.  So $\mathscr{G}_x$ is much, much larger than $\mathscr{F}_x$.
Where is the mistake?
Update: Thanks to both of you for helpful answers; it was tough choosing just one.
 A: First let me note that the asociation  $\mathcal F\to\mathcal G=\mathcal G(\mathcal F)$ is an extremely useful functor introduced by Godement in his book and through which he calculates the Grothedieck  cohomology of sheaves, the main virtue of his sheaves being that they are flabby, hence acyclic.
So your choice of the letter $\mathcal G$ for this Godementification is excellent!
And you are perfectly right that the stalk $\mathcal G_x$ is much, much larger than the corresponding $\mathcal F_x$: congratulations for not sharing the common misconception that these stalks are equal.     
Your difficulty just results from two facts:   
1) Your notation $i_\ast \mathcal F_x$ does not make sense: what is $i$?
You might introduce  the collection of $i_x$'s (and then take into account Matt's remark)  but it is better to directly define $\mathcal G$ by your (and Godement's!)  formula $\mathcal{G}(U) :=  \prod_{x \in U} \mathcal{F}_x$.   
2) More importantly, it is completely false that taking stalks commutes with infinite products.
This explains the discrepancy between your two reasonings.
A: In the third line you write that taking stalks commutes with direct sums, but the actual construction of $\mathcal G$ is a direct product, not a direct sum (unless $X$ is finite).  
Even if $X$ is finite, it won't be true that $\mathcal F = \mathcal G$ unless all the points $x$ are closed (so that $X$ is actually discrete, in which case
there is no difference between continuous and discontinuous sections).
The reason is that if $x$ is not closed, then $i_*(\mathcal F_x)$ has non-zero stalks not just at $x$, but at all the points in the closure of $x$.
