I am studying the first page of this article here.
The article defines a differential $1$-form to be a smooth map $\alpha : TM \to \mathbb R$ ($TM$ here is the tangent bundle) such that for $m \in M$ the point $\alpha (m)$ is a linear functional $T_m M \to \mathbb R$.
But then shortly afterwards the author writes that the $1$-forms $dx_i$ are the $i$-th projection: $dx_i (v) = v_i$.
But these $dx_i$ do not appear to be evaluated at any point $m \in M$. Isn't this contradicting the definition given earlier?
Please could someone help me understand what's going on here?
If you ever want to get anywhere in geometry, you are going to have to learn to be more flexible with your notational qualms. I wish this were not the case, but this is the culture of the field.
Here, in order to make sense of $dx_i(v)$, you must first realize that $v$ is supposed to be a vector, in order to make sense of projection. Since it is standing in for an element of the tangent bundle, it is not a vector. But it is contained in some fibre of the manifold, so it can be reinterpreted as a vector in the tangent space at some point $m$. So $dx_i(v)=v_i$ really means $dx_i((m,v))=v_i$, but the $m$-dependence is suppressed because $m$ depends completely on $v$.
[Of course, all of this is happening with respect to coordinates! And these coordinates are not even uniquely defined, so you must either slog through a well-definedness argument or believe that $dx_i$ is even a thing that makes sense.]