In both Algebraic Geometry by Hartshorne and Geometry of Schemes by Eisenbud and Harris, the authors describe the notion of a family of schemes as being the fibres of a morphism $f:X\to Y$. Or as Eisenbud/Harris put it $X$ can be loosely thought of as a family of schemes parametrized by the points in $Y$.
I'm having a hard time seeing why this definition gives us the notion of parametrization. One would expect a parameter to be a member of the domain. But instead we have the parameter lying in the image scheme.
Consider a the (set-theoretic) function $f:X\to Y$ and let us understand what it means for this to be a "family" of stuff. As you mentioned each point in $Y$ should parametrize something for the notion of family to be meaningful.
Define $\Phi_y = f^{-1}(y)$. Now as a set $X=\bigcup_{y\in Y}\Phi_y$. The parameterization you have in mind (intuitively) is exactly the assignment $y\mapsto \Phi_y$. It assigns to each point of $y$ some set $\Phi_y$. So the map $f:X\to Y$ gives you a parameterization.
Conversely suppose you have a family of objects $\{F_b\}_{b\in B}$, parametrized by some set $B$. Crudely stack them on top of each other to obtain $E=\coprod_{b\in Z}F_b$ and define $f: E\to B$ by $f(b,e)=b$ with $b\in B$ and $e\in F_b$. This gives you the desired function $f:E\to B$.
Of course the above construction is very crude and most of the time useless unless there are some conditions on $f:X\to Y$. The least amount of conditions you will see is $X,Y$ be topological spaces and $f:X\to Y$ be a bundle, for a "good" family. But "good" family is different depending on what you want exactly.
In your example of $t\mapsto \mathbf{v}t+(1-t)\mathbf{w}$, a parametrization by $I$ (unit interval) of points on a segment $S$ between two vectors $\mathbf{v}, \mathbf{w}$, the function $f$ is defined as follows: $f:S\to I$ given by $f(\mathbf{v}t+(1-t)\mathbf{w})=t$.
Perhaps a better, but still simple, example would be the projective space $\mathbb{P}^n$ over $\mathbb{C}$ (or $\mathbb{R}$ doesn't matter). Each point of $\mathbb{P}^n$, say $x$, parametrizes a lines passing through the origin and $x$ (regarded as non-projective coordinates) in $\mathbb{A}^{n+1}$. Now the natural map $$ \pi: \mathbb{A}^{n+1}\setminus\{0\}\to \mathbb{P}^n $$ is exactly this family of lines (in the modern terminology) since the fibers $\pi^{-1}(x)=\{\lambda x\in \mathbb{A}^{n+1}\setminus \{0\}\mid \lambda\neq 0\}$ are exactly that line that passes through the origin and $x$, with the origin removed. If the removal of the origin bothers you, it can be fixed via the blowup of $\mathbb{A}^{n+1}$ at the origin (if you don't know blowup look up Hartshorne I.4 for example). If we denote this blowup by $\tilde{\mathbb{A}}^{n+1}\subset \mathbb{A}^{n+1}\times \mathbb{P}^n$, then using $\pi$, one can define a natural map $$ \pi': \tilde{\mathbb{A}}^{n+1}\to \mathbb{P}^n $$ with fibers now exactly the whole lines passing through the origin in $\mathbb{A}^{n+1}$ (including the origin).
As a side note there is also a natural blowup map $\varphi: \tilde{\mathbb{A}}^{n+1}\to \mathbb{A}^n$ such that $\varphi^{-1}(0)\simeq \mathbb{P}^n$ and $\tilde{\mathbb{A}}^{n+1}\setminus \varphi^{-1}(0)\simeq \mathbb{A}^{n+1}\setminus \{0\}$. Intuitively for each point of $\mathbb{P}^n$ we need a copy of the origin if we are to stack these line on top of each other. This is exactly what $\varphi^{-1}(0)\simeq \mathbb{P}^n$ does. The reason behind the name "blowup" at origin and also the construction of the map $\pi'$ should be clear now... For example $\tilde{\mathbb{A}}^2$ looks like an (infinite) Mobius band.