$\newcommand{\dd}{\partial}$In the sense that you're asking, velocity of a point particle is a vector, and any object that "measures a velocity and returns a scalar" is a covector.
Though there's a meaningful mathematical distinction between a vector space $V$ and its dual space $V^{*}$, it's probably easier to see the distinction between a vector-valued function and a covector-valued function, because (as lists of component functions) they transform differently under change of coordinates.
For simplicity, assume we're in a plane region, equipped with coordinates $(x_{1}, x_{2})$. Rather than writing vectors as column matrices and covectors as row matrices, introduce the coordinate vector fields
$$
\frac{\dd}{\dd x_{1}},\qquad
\frac{\dd}{\dd x_{2}},
\tag{1a}
$$
and the coordinate one-forms
$$
dx_{1},\qquad
dx_{2}.
\tag{1b}
$$
A smooth vector field is a "linear combination" of the coordinate fields with smooth functions coefficients, e.g.
$$
X = X^{1}\, \frac{\dd}{\dd x_{1}} + X^{2}\, \frac{\dd}{\dd x_{2}}.
\tag{1c}
$$
A smooth covector field (or smooth one-form) is a "linear combination" of the coordinate one-forms with smooth functions coefficients, e.g.
$$
\xi = \xi_{1}\, dx_{1} + \xi_{2}\, dx_{2}.
\tag{1d}
$$
If $(y_{1}, y_{2})$ is another coordinate system, the chain rule gives
\begin{align*}
\frac{\dd}{\dd x_{1}}
&= \frac{\dd y_{1}}{\dd x_{1}}\, \frac{\dd}{\dd y_{1}}
+ \frac{\dd y_{2}}{\dd x_{1}}\, \frac{\dd}{\dd y_{2}}, &
\frac{\dd}{\dd x_{2}}
&= \frac{\dd y_{1}}{\dd x_{2}}\, \frac{\dd}{\dd y_{1}}
+ \frac{\dd y_{2}}{\dd x_{2}}\, \frac{\dd}{\dd y_{2}};
\tag{2a} \\
dx_{1}
&= \frac{\dd x_{1}}{\dd y_{1}}\, dy_{1}
+ \frac{\dd x_{1}}{\dd y_{2}}\, dy_{2}, &
dx_{2}
&= \frac{\dd x_{2}}{\dd y_{1}}\, dy_{1}
+ \frac{\dd x_{2}}{\dd y_{2}}\, dy_{2}.
\tag{2b}
\end{align*}
Writing
$$
X = Y^{1}\, \frac{\dd}{\dd y_{1}} + Y^{2}\, \frac{\dd}{\dd y_{2}},
$$
substituting (2a) into (1c), and equating components gives the transformation rule for vector fields:
$$
Y^{1}
= X^{1}\, \frac{\dd y_{1}}{\dd x_{1}}
+ X^{2}\, \frac{\dd y_{1}}{\dd x_{2}},\quad
Y^{2}
= X^{1}\, \frac{\dd y_{2}}{\dd x_{1}}
+ X^{2}\, \frac{\dd y_{2}}{\dd x_{2}}.
\tag{3a}
$$
Similarly, writing
$$
\xi = \eta_{1}\, dy_{1} + \eta_{2}\, dy_{2},
$$
substituting (2b) into (1d), and equating components gives the transformation rule for one-forms:
$$
\eta_{1}
= \xi_{1}\, \frac{\dd x_{1}}{\dd y_{1}}
+ \xi_{2}\, \frac{\dd x_{2}}{\dd y_{1}},\qquad
\eta_{2}
= \xi_{1}\, \frac{\dd x_{1}}{\dd y_{2}}
+ \xi_{2}\, \frac{\dd x_{2}}{\dd y_{2}}.
\tag{3b}
$$
Classically, a vector field is a collection of functions associated to a coordinate system that transform like (3a), and a one-form is a collection of functions that transforms like (3b). And that, in a sense, is the distinction between vectors and covectors.
Admittedly, the coordinate vector field and coordinate one-form notation is merely a formalism, just like column and row matrices. The tie-in with the chain rule, however, may make this formalism appealing, even compelling.
The connection with matrix notation comes from expressing (3a) and (3b) in terms of matrix products. For brevity, write
$$
a_{j}^{i} = \frac{\dd y_{i}}{\dd x_{j}},\qquad
b_{j}^{i} = \frac{\dd x_{i}}{\dd y_{j}},
$$
so that the matrices
$$
A = \left[\begin{array}{@{}cc@{}}
a_{1}^{1} & a_{2}^{1} \\
a_{1}^{2} & a_{2}^{2} \\
\end{array}\right] = \frac{\dd(y_{1}, y_{2})}{\dd(x_{1}, x_{2})}
$$
and
$$
B = \left[\begin{array}{@{}cc@{}}
b_{1}^{1} & b_{2}^{1} \\
b_{1}^{2} & b_{2}^{2} \\
\end{array}\right] = \frac{\dd(x_{1}, x_{2})}{\dd(y_{1}, y_{2})},
$$
the Jacobian of the change of coordinates from $x$ to $y$ and the Jacobian from $y$ to $x$, are inverses.
If we identify
$$
\frac{\dd}{\dd x_{1}} = \left[\begin{array}{@{}c@{}}
1 \\
0 \\
\end{array}\right],\qquad
\frac{\dd}{\dd x_{2}} = \left[\begin{array}{@{}c@{}}
0 \\
1 \\
\end{array}\right],\qquad
X = \left[\begin{array}{@{}c@{}}
X^{1} \\
X^{2} \\
\end{array}\right],
$$
and
$$
dx_{1} = \left[\begin{array}{@{}cc@{}}
1 & 0 \\
\end{array}\right],\qquad
dx_{2} = \left[\begin{array}{@{}cc@{}}
0 & 1 \\
\end{array}\right],\qquad
\xi = \left[\begin{array}{@{}cc@{}}
\xi_{1} & \xi_{2} \\
\end{array}\right],
$$
then (3a) and (3b) read, respectively,
$$
\left[\begin{array}{@{}c@{}}
Y^{1} \\
Y^{2} \\
\end{array}\right]
= A \left[\begin{array}{@{}c@{}}
X^{1} \\
X^{2} \\
\end{array}\right]
$$
and
$$
\left[\begin{array}{@{}cc@{}}
\eta_{1} & \eta_{2} \\
\end{array}\right]
= \left[\begin{array}{@{}cc@{}}
\xi_{1} & \xi_{2} \\
\end{array}\right] B.
$$
Particularly,
$$
\xi(X) = \xi_{1} X^{1} + \xi_{2} X^{2} = \eta_{1} Y^{1} + \eta_{2} Y^{2};
$$
the pairing between a one-form and a vector field is independent of the coordinate system, and consequently represents a "physical" or "geometric" concept.
Best Answer
Given a vector space $V$, there is a "dual" space $V^*$ which consists of linear functions $V\to \mathbb{F}$ (where $\mathbb{F}$ is the underlying field). Given $v\in V, \phi \in V^*$, we can plug in to get a number $\phi(v)$.
Because of linearity, $V^*$ is actually a vector space. If $V$ is finite dimensional, then $V^*$ has the same dimension. One way to see this is, if we fix a basis $e_1, \ldots e_n \in V$, we have a basis $\phi_i, \ldots, \phi_n \in V^*$ defined by $\phi_i(e_j)=\delta_{ij}$, which is $1$ when $i=j$ and $0$ otherwise.
Of course, this isomorphism requires choosing a basis, and in general, there is no "natural" choice of isomorphism. Additionally, when $V$ is infinite dimensional, $V$ and $V^*$ will not be isomorphic.
So if we are just doing basic linear algebra, there is no real difference between vectors and covectors. There are some constructions that might seem to require a choice of basis if you don't use covectors (like taking the transpose of a matrix), but they are not fundamentally different kinds of objects. However, if we want to work geometrically, we can see a difference.
Given a manifold $M$, and a point $p\in M$, we have a vector space $T_pM$ of the tangent vectors to $M$ at $p$. For example, if you take the hollow sphere sitting inside $R^3$, you can look at the plane that sits tangent to a point, and turn it into a vector space. These tangent vectors act on functions by taking the directional derivative of a function at a point. If you take a tangent covector, it no longer acts on functions, it just acts on vectors. Geometrically speaking, it is a fundamentally different kind of object. By taking a tangent vector at every point, you get something called a vector field, but taking a covector at every point you get something called a differential form. They are both useful notions, but they are used in fundamentally different ways.
Of course, once you get the general notion of a vector bundle (essentially, a way of smoothly putting a vector space at every point of a manifold), you can see that tangent vectors and tangent covectors are just dual vector bundles, and in the absence of certain geometric constructions can be treated very similarly.