Scheme over a field vs. scheme defined over a field A scheme $X$ over another scheme $S$ is simply a morphism $X \to S$. In particular, if $K$ is a field, a scheme $X$ over $K$ is given by a morphism $X \to \operatorname{Spec} K$. For a long time, I conflated this with the notion of being defined over $K$.
However, the morphisms don’t even go in the right direction: For if $k \subseteq K$ is a subfield, we have a morphism $\operatorname{Spec} K \to \operatorname{Spec} k$ and composing it with our morphisms $X \to \operatorname{Spec} K$ above allows us to consider $X$ as a scheme over $k$ as well, but $X$ should not automatically be defined over every subfield (which intuitively I think of as being given by polynomial equations with coefficients in $k$).
So here is where I am at: The question “Is (a general scheme) $X$ defined over a field $k$?” does not make sense. We can only ask it if $X$ is already given as a scheme over some extension $K$ of $k$. In this case, we say that $X$ is defined over $k$ if there is a scheme $X_k$ over $k$ such that the fiber product $X_k \times_k \operatorname{Spec} K$ is $X$ (or rather, isomorphic to $X$ as a scheme over $k$).
In particular this would entail that a scheme over $k$ is always defined over $k$ (as a scheme over $k$!). This might explain my long-time confusion.
Is this understanding correct?
 A: Here $k$ and $K$ are fields, and a "$k$-scheme" just means a scheme over $k$.
That's right, you don't say that a $K$-scheme $X$ is defined over a subfield $k \subset K$ just by composing the morphisms $X \rightarrow \operatorname{Spec} K \rightarrow \operatorname{Spec} k$.  It's more subtle than that.  In particular, one reason we don't do this is because for example if $K = \overline{k}$ then $X$ would no longer be of finite type over $k$ even if it is so over $K$.  Usually the kind of $k$-schemes people are interested in are varieties, which by any reasonable definintion (there are some minor differences between different authors in the definition of variety) are finite type.
The usual context in which you encounter the language you're talking about is when you have an ambient $k$-variety $\mathscr X$, and $X$ is a subscheme of the $K$-scheme $\mathscr X_K = \mathscr X \times_{\operatorname{Spec}(k)} \operatorname{Spec}(K)$.  By composition $X \rightarrow \mathscr X_K \rightarrow \operatorname{Spec}(K)$, $X$ is naturally a $K$-scheme.  We say $X$ is defined over $k$ if there is a subscheme $X_0$ of $\mathscr X$ such that $(X_0)_K = X$.  If $X_0$ exists, then it is unique.
A: I don't know what you mean by "being defined over a field" but this notion is naturally generalised, in the language of schemes, with the concept of relative schemes that you mention. 
Perhaps you mean restriction and extension of scalars? 
Given a $K$-scheme $X$, for every extension $L\supset K$ we can construct the scheme $X_L:= X\times_K \mathrm{Spec}(L)$ where the fibre product is taken between the morphisms $X\longrightarrow \mathrm{Spec}(K)$ (the structure morphism of $X$) and $\mathrm{Spec}(L)\longrightarrow \mathrm{Spec}(K)$ (the morphism induced by inclusion $K\hookrightarrow L$). This scheme $X_L$ is extension of scalars (or base change) and is naturally a $L$-scheme together with the natural projection associated to the fibre products.
Geometrically, if your scheme $X$ is defined locally by equations in $k[x_0,\ldots,x_n]$, the scheme $X_L$ will be defined by the same equations viewed in $L[x_0,\ldots,x_n]$.
If you want to restrict scalars from $K$ to $k\subset K$, things are not so easy. However, this is possible, it the unique object that represents the Weil restriction functor. It can be seen also as right adjoint to the fibre product so the construction is compatible with extension of scalars.
When I say this is not an easy operation I mean that the restriction functor does not preserve many geometric propertis like the extension functor does.
