Do 1-forms return scalars or co-vectors? I've been browsing questions regarding 1-forms and their difference with co-vectors, and I have stumbled upon what follows.
@magma, here, said:

1-forms are simply the linear operators that take a vector and give
out a number

So, thinking of a function, the range of a 1-form is a real value.
On the other hand, @Silly Goose, here, said

... if you have a 1-form α on some manifold M and you pick any x∈M,
the value of α at x is a co-vector.

Doesn't the latter quote assert that a 1-form, functionally, returns a co-vector, thereby contradicting the former quote?
 A: Remember that you need to put in two inputs into a 1-form before you get a scalar: a point on your manifold, and then a vector at the tangent space of that point. If you fix the point $x\in M$ but not a vector $v\in T_xM$, then you have a covector $\omega|_x:T_xM\to \mathbb{R}$, i.e. $\omega|_x\in T_x^*M$. If you fix a vector field $X$ on $M$ but not a point, then you have a function $\omega(X):M\to \mathbb{R}$. If you fix neither, then you have the usual definition of a 1-form, namely
$$\omega:M\times T_x M\to \mathbb{R}.$$
The abstract thing that's happening is if you have a map $X\times Y\to Z$, you can view this same map in three ways. First, exactly as written. Second, if you fix an $x\in X$, this becomes a map $Y\to Z$. Third, if you fix a $y\in Y$, then this becomes a map $X\to Z$.
A: These are subtly different.  All one forms on a given (assume finite dimensional vector space) are linear maps from vectors to the field.  On a manifold each point has a vector space associated with it (the tangent space $T_pM$) upon which one can define one forms as above.  The collection of all one forms is commonly denoted as $T_pM^*$. The difference here is that now a one form on a manifold is a function:
$$\omega:M \to T_pM^*$$
And the properties must interact in a "nice" way with the smooth structure of $M$ or if you like the tangent bundle $TM$.
A: The reconciling difference is in one word that is missing in the latter quote: differential 1-form. Differential 1-form is indeed a smooth field of (algebraic) 1-forms. This is described in @Nick's answer
