I'll use the standard notation for the ultraproduct of $(X_i)_{i \in I}$ with respect to an ultrafilter $\mathcal{U}$ on $I$: $$\Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U}$$ (for either of the two definitions). The notation makes more sense for type 1 than type 2, but never mind.
(1) Ultraproduct with respect to a principal ultrafilter is projection. That is, if $k \in I$ and $\mathcal{U}$ is the principal ultrafilter on $k$ then $(\prod X_i)/\mathcal{U} = X_k$. This is true without exception for type 2 ultraproducts. It is almost true for type 1, but fails if $X_k \neq \emptyset$ and $X_i = \emptyset$ for some $i \neq k$.
(2) Ultraproducts preserve finite coproducts. That is, writing $+$ for the coproduct (disjoint union) of sets, $$\Bigl( \prod_{i \in I} (X_i + Y_i) \Bigr) \bigl/ \mathcal{U} \cong \Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U} + \Bigl( \prod_{i \in I} Y_i \Bigr) \bigl/ \mathcal{U}$$ for any families of sets $(X_i)$ and $(Y_i)$. This is true without exception for type 2 ultraproducts (using the fact that these are ultra$\mbox{}$filters). But again, it isn't quite true for type 1: it can fail when some of the sets are empty. For example, let $I$ be a set, partition it into two choose any nonempty subsets proper subset $J$ and of $K$, I$, and put $$X_i = \begin{cases} 1 &\text{if } i \in J\\ \emptyset &\text{otherwise,} &\text{otherwise} \end{cases} \qquad\ \qquad Y_i = \begin{cases} 1 \emptyset &\text{if } i \in K\\ J\\ emptyset 1 &\text{otherwise,} \end{cases}$$ where$1$denotes a one-element set. Then according to the type 1 definition,$(\prod(X_i + Y_i))/\mathcal{U} = 1$but$(\prod X_i)/\mathcal{U} + (\prod Y_i)/\mathcal{U} = \emptyset$. I guess all of these things that go wrong are intimately related to Łoś's theorem, which Michael alluded to. 1 Michael Barr pointed out one thing that goes wrong if you try to use the type 1 definition when some of the sets involved are empty. Now that I understand this issue better, I'll point out a couple of other things that go wrong. They're both of the form "this theorem holds cleanly for type 2 ultraproducts, but for type 1 you have to make exceptions". I'll use the standard notation for the ultraproduct of $(X_i)_{i \in I}$ with respect to an ultrafilter$\mathcal{U}$on$I$: $$\Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U}$$ (for either of the two definitions). The notation makes more sense for type 1 than type 2, but never mind. (1) Ultraproduct with respect to a principal ultrafilter is projection. That is, if$k \in I$and$\mathcal{U}$is the principal ultrafilter on$k$then$(\prod X_i)/\mathcal{U} = X_k$. This is true without exception for type 2 ultraproducts. It is almost true for type 1, but fails if$X_k \neq \emptyset$and$X_i = \emptyset$for some$i \neq k$. (2) Ultraproducts preserve finite coproducts. That is, writing$+$for the coproduct (disjoint union) of sets, $$\Bigl( \prod_{i \in I} (X_i + Y_i) \Bigr) \bigl/ \mathcal{U} \cong \Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U} + \Bigl( \prod_{i \in I} Y_i \Bigr) \bigl/ \mathcal{U}$$ for any families of sets$(X_i)$and$(Y_i)$. This is true without exception for type 2 ultraproducts (using the fact that these are ultra$\mbox{}$filters). But again, it isn't quite true for type 1: it can fail when some of the sets are empty. For example, let$I$be a set, partition it into two nonempty subsets$J$and$K$, and put $$X_i = \begin{cases} 1 &\text{if } i \in J\\ \emptyset &\text{otherwise,} \end{cases} \qquad\ \qquad Y_i = \begin{cases} 1 &\text{if } i \in K\\ \emptyset &\text{otherwise,} \end{cases}$$ where$1$denotes a one-element set. Then according to the type 1 definition,$(\prod(X_i + Y_i))/\mathcal{U} = 1$but$(\prod X_i)/\mathcal{U} + (\prod Y_i)/\mathcal{U} = \emptyset\$.