The graphs of degree one polynomial functions are the same as non-vertical lines. Vectors are not lines; as you indicate, they consist of direction and (finite) length (and, depending on context, a base point).
There is a close relationship between the two. A line is determined by taking a point in the plane and then choosing the direction to travel from that point. This direction can be specified by a vector. So instead of thinking of a line as the graph of a degree one polynomial function you could think of it as the set of all points $p+tv$, where $p$ is point in the plane, $v$ is a nonzero vector, and $t$ ranges over the real numbers. The parameter $t$ tells you how much you travel in the $v$ direction (technically, divided by the length of $v$), with the understanding that when $t$ is negative you are going $|t|$ much in the opposite direction. This is actually more general than thinking of graphs of functions, because you could also describe vertical lines by choosing $v$ to point straight up or down.
To summarize: A vector and a starting point determine a line, and a line determines a direction vector up to multiplication by a nonzero real number.
To be concrete, consider $f(x)=2x+3$. The point $p=(0,3)$ is on the graph, and the slope $2$ determines the direction. This direction can be described by the vector $v$ with $x$ component $1$ and $y$ component $2$. We can write $v=(1,2)$ to summarize this information. Here going $t$ in the $v$ direction from $p$ takes you to the point $p+tv=(0,3)+(t,2t)=(t,2t+3)$, which is the same as the point $(t,f(t))$ on the graph of $f$. Notice that the slope of the line is the ratio of the $y$ component to the $x$ component of the vector.
On the other hand, say we take the point $p=(3,-1)$ and the vector $v=(-3,6)$. The line through $p$ in the $v$ direction consists of all of the points $p+tv=(3,-1)+(-3t,6t)=(3-3t,6t-1)$ as $t$ ranges over the real numbers. To determine the corresponding function, just find the slope-intercept form of the equation of the line through the point $(3,-1)$ with slope $\frac{6}{-3}=-2$, namely $y=-2x+5$, so that $f(x)=-2x+5$ is the desired function.
