This is going to be perhaps vague, but I'l try to write down the idea.
I've been told that in doing scheme theory, most of the time what is studied is not a scheme but a morphism of schemes $f:X\rightarrow S$. The studied of such a maps can be enriched allowing a chance of "base" $S$. Such a scheme $S$ can be crazy, but I'll impose finite properties to the map $f$ to keep everything under control.
Now, as you say, if we have two $S$-schemes $f:X\rightarrow S$ and $Y\rightarrow S$ then we can consider the fiber product of them (which is a categorical product of $X$ and $Y$ over $S$). Such a process can be though of as replacing the base $S$ by the scheme $Y$. In doing so, the new map $f_Y:X_Y\rightarrow Y$ (standing for $p_2:X\times_{S}Y\rightarrow Y$) may be easier to work with than the map itself $f$. This very construction is generalizing the idea of "extending scalars".
Here is a sort of example of a product described above. Given the point $s\in S$ and its residue field $k(s)$ of the local ring $\mathcal{O}_s$ at that point. Then it is well known that $X_s=X\times_S Spec(k(s))$ has an underlying space the "fiber" $f^{-1}(s)$ (as long as $f$ is "finite"). This space can be considered as an algebraic variety over the field $k(s)$. This way $S$ is the scheme parameterizing the varieties $X_s$ some of which a priori may have different ground fields $k(s)$.
On the other hand, back to the fiber product, in considering a variety $X$ over the field $k$ (scheme of finite type over $k$), we can extend the scalars from $k$ to a field extension $K$. In doing so, we may think of such a variety now over a field extension $K$. Here is where the product comes into play. Vaguely enough, this is saying that if you have an equation and you have solved it over the field $k$ now you might ask yourself if you'd be able to solved it again but over a field extension $K$, here you want to perform an extension of scalars and the ideas written above may help out.
Needless to say that we can consider as well $X\times_k Spec(\overline{k})$ and due to the fact that field $k$ may not have been algebraically closed, such a fiber product can be handful.
Now, to get intuition, let's take a look at an example: $$\pi:Spec \mathbb{Z}[i]\rightarrow Spec(\mathbb{Z})$$
Let $X=Spec\mathbb{Z}[i]$ and notice that the fiber $$X_p=X\times_{\mathbb{Z}}\mathbb{F}_p=Spec(\mathbb{Z}[i]\otimes\mathbb{F}_p)$$
Such a fiber is going to have cardinality 2 if $p\equiv 1 mod(4)$ and cardinality 1 if $p\equiv 3 mod(4)$ (this is the fact that $p=a^2+b^2$ where $a,b\in \mathbb{Z}$ if $p\equiv 1mod 4$). These fibers are $X$ reduce mod $p$. Notice that at the generic point we have $X_0=\mathbb{Q}[i]$.
Now the $T-points$ in $X$ where $T=\mathbb{F}_p[i]$, and all that story give rise to the functor which represents the scheme $X$. If I am not wrong, the info that carries such a functor is nothing but that of the fibers of the map $\pi$. So naively it is like having $X$ fibers wise.
(1) A family of hypersurfaces of degree $d$ over an affine scheme $Spec\ R$ means the following: it is a closed subscheme of $\mathbb P^n_R$ given by a homogeneous polynomial $f(x_0,\dots,x_n)$ of degree $d$ satisfying the following condition:
Every fiber is a hypersurface of degree $d$. This means that for every prime ideal $m\subset R$, the reduction $\bar f\in k[x_0,\dots,x_n]$ is a polynomial of degree $d$, where $k=R/m$.
Now, let $m$ be a maximal ideal, so that $k$ is a field, and look at the graded ring $k[x_0,\dots,x_n]/(f)$. For each $a\ge d$, the degree-$a$ part is a vector space of dimension $\binom{a+n}{n}-\binom{a-d+n}{n}$. Thus, some $\binom{a-d+n}{n}$ monomials can be written as linear combinations, with coefficients in $k$, of the remaining monomials. This is done by solving a system of linear equations obtained by setting $x^m f=0$, where $x^m$ are monomials of degree $a-d$.
Now, consider the same system of linear equations with coefficients in $R$. For each of the $\binom{a-d+n}{n}$ monomials as above you get a principal minor $M$ of your matrix, and the reduction of $\det M$ in $k$ is not zero. Thus, over an open set $Spec\ R[1/\det M]$, this determinant is invertible, and the monomial can be eliminated.
Thus, over an open neighborhood $Spec\ A$ of the point $[m]\in Spec\ R$, the degree-$a$ part of the ring $A[x_0,\dots,x_n]/(f)$ is a free $A$-module. Recalling how $Proj$ is covered by $Spec$'s and that a free module is flat, this implies that $Proj\ S[x_0,\dots,x_n]/(f)$ is flat over $Spec\ S$. (We will assume $R$ and so $A$ to be Noetherian here for simplicity.)
This is the main trick for proving flatness over a non-reduced base: you prove freeness instead. For a finitely generated module over a Noetherian ring, flatness and freeness are equivalent. So for a projective morphism $f:X\to Y$ a coherent sheaf $F$ on $X$ is flat over $Y$ iff the sheaves $f_* F(a)$ on $X$ are locally free for $a\gg0$.
In (2), you are mistaken about Hartshorne: Theorem III.12.11 (Cohomology and Base Change) has no assumption for the base to be reduced. So if $H^i(X_y,F_y)=0$ for $i>0$ then $f_*F$ is locally free, for any (Noetherian) base $Y$ and coherent sheaf $F$ on $X$, flat over $Y$.
For higher direct images, $H^i(X_y,F_y)=0$ for $i\ge i_0$ implies that $R^{i_0}f_*F=0$. But you can have $H^i(X_y,F_y)=0$ for $i> i_0$ and $H^{i_0}(X_y,F_y)$ non-constant, and still have $R^{i_0}f_*F=0$ (compare the Poincare line bundle on $A\times A^t$, as in Fourier-Mukai).
Best Answer
I think if you want a good answer to your question you will have to consult papers by people who don't think F_1 is a scam! To be more precise: since F_1 doesn't actually exist, what do people think it is? And I think most people would say words like: there should be some category of schemes over Z which are somehow "isotrivial" or "the same over every prime," and we call this category the category of schemes over F_1, and this serves as a "definition" of F_1. And, in addition, its supplies the name you requested for schemes with this good behavior.
I think it is stronger than having all cohomology of Tate type, though I'm no expert on this. Grassmannianns are schemes over F_1 (I'm pretty sure) but elliptic curves never are.
Now this is all just philosophy unless without an actual definition somewhere in the picture. The good news is, there is one! Actually, several! The one I like best is due to Jim Borger:
http://arxiv.org/abs/0906.3146
So have a look at this and see if "schemes with Lambda-structure" are what you're looking for.