Neither of these claims are true, and the reason is essentially that "The topology on $X\times Y$ is not the product topology". This is more or less the claim you're making in part $2)$, and then this failure causes $1)$ to fail. The fact that the topology isn't just the product topology is maybe not entirely surprising at the level of schemes (since the product itself isn't even the product as sets), but I think it is relatively surprising if you're thinking about this on the level of classical varieties, where you can just define the set of the product as the usual product, but then have to work harder to define the topology. This difference is because one of the main advantages of using schemes rather than classical varieties is that the points of a scheme contain a lot of information about the topology then the points of the classical variety (which are just the closed points of the scheme).
An example where $2)$ fails is where $X = Y = \mathbb{A}^1_k$, with $k = \bar{k}$ for simplicity. Then any non-empty open set in $\mathbb{A}^1_k$ is just the complement of a finite set of (closed) points. Explicitly, any open set is of the form $D(f)$ for some $f$ and corresponds to the complement of the roots of $f$. Then the product $D(f) \times_k D(g) = \pi_X^{-1}(D(f)) \cap \pi_Y^{-1}(D(g))$ corresponds to the complement of a finite union of lines in $\mathbb{A}^2_k$, corresponding to vertical lines corresponding to the roots of $f$ and horizontal lines corresponding to the roots of $g$. Note that a union of sets of this form is still of this form, so the topology generated by these sets is simply their union. In particular, there are many open sets not of is form, e.g. the diagonal $D(X-Y)$, so the actual topology is much richer than the product topology.
Moreover, $1)$ is also not true. You could probably just compute the product of the preimage sheaves in the example above, but there is an easier way that fits in with the philosophy used above that the product of schemes is more than just the naïve product. As well as the topology not being the product topology, the set is not even the product of the sets! In particular, there may be distinct points in the product that project to the same points in $X$ and $Y$. To see that this is a problem, note that:
$$(\pi^{-1}_X \mathcal{O}_X \otimes \pi^{-1}_Y\mathcal{O}_Y)_z \cong (\pi^{-1}_X \mathcal{O}_X)_z \otimes_k (\pi^{-1}_Y\mathcal{O}_Y)_z \cong \mathcal{O}_{X, \pi_X(z)} \otimes_k \mathcal{O}_{Y,\pi_Y(z)} $$
And so in particular, points with the same projections to $X$ and $Y$ should have the same stalks. However, with our example above, notice that the primes $(X-Y)$ and $(0)$ both map to the generic point under either projection to $\mathbb{A}^1$, and so the stalks of the product sheaf at both points are the same. However, the stalks of the structure sheaf are different (and we can even see this without calculation, the local rings have different dimensions since the closures of the two points have different codimension in $\mathbb{A}^2_k$).
As Paf points out in the comments to your question, you can look at this post on mathoverflow for a little more discussion on what the points of the fibre product are, and what stalks of the structure sheaf of the product are at said points. In particular there are a wealth of points that where the product sheaf will have the same stalks, but the structure sheaf won't. Notice that the same argument will extend to show that we can't even replace the tensor product of the preimage sheaves with the tensor product of the pullback sheaves, or any other reasonable extension.
Best Answer
The structure sheaf is actually not so simple. On base open affines $U_f$, we define it to be $\mathcal{O}_X(U_f) = A_f$ (localization), but as you say this doesn't tell us what $\mathcal{O}_X(U)$ should be if $U$ is open but not affine.
If you look at Hartshorne, for example, you'll see that the better way to proceed is to first define the stalks of $\mathcal{O}_X$, and then take $\mathcal{O}_X(U)$ to be the ring of local sections (maps from $U$ to the stalks over points in $U$ satisfying some nice conditions). It is not too hard to check that on an open affine $U_f$, this is equivalent to $\mathcal{O}_X(U_f) = A_f$. This is very similar to sheafification of a presheaf.
Another way to think of this, if it helps, is that since any open set $U$ can be covered by affine open subsets $\{U_{f_i}\}_{i \in I}$, so a section of $\mathcal{O}_X(U)$ can be thought of as a collection of sections $s_i \in A_{f_i}$, satisfying the condition that $s_i$ and $s_j$ agree on the intersection of $U_{f_i}$ and $U_{f_j}$ (and we can cover $U_{f_i} \cap U_{f_j}$ by open affines if necessary, to make sense of this condition). The point is that since the base open affines $U_f$ form a base of the topology on $X$, any sheaf on $X$ is uniquely determined by its values on these base open affines.
There was a good MO thread on this, which can be found here.