Here is a how a typical proof might look like in group theory--- Suppose we are given a finite group $G$. Enumerate the elements $g_1, \dots, g_n$. Now consider a formula $\phi(g_1, \dots, g_n)$ which discusses some property of $G$....
This proof is not rigorous. In any set theory I can think of, a finite set is equinumerous to $\{0, \dots, n}$ where $n$ may be a non-standard integer. So when we are proving things about finite sets, we really should allow the size of the set to be non-standard. But, at least implicitly, any proof assumes that $n$ is a standard integer.
This seems to be different than proofs of arithmetic, in which we can assume that the computations are occurring in the true structure of $\mathbf N$, which contains only standard integers. We may need extra axioms to pin down the behavior of $\mathbf N$, but we ultimately will never need the non-standard integers.
EDIT: I think I have boiled down this issue to the the following example.
We will prove that $CON(ZFC) \vdash (ZFC \vdash CON(ZFC))$, which is obviously false.
Consider the following proof. Suppose CON(ZFC) is true. Given a model $M$ of ZFC and $n$ an integer of $M$. If $n$ is a standard integer, then by assumption $n$ is not a proof of contradiction of ZFC. So we have shown that for all standard integers $n$ and all models $M$ of ZFC, $n \in M$ does not disprove ZFC. Hence, by the transfer principle, ZFC proves that all integers $n$ are not disproofs of ZFC. Hence ZFC proves CON(ZFC).
The key issue is that some formulas do behave differently for standard and non-standard formulas. In this case, the formula "$x$ is a disproof of ZFC" is a formula which picks out the non-standard integers.
