From physics classes, I understand the electromagnetic field strength tensor to be defined as $$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu \;,$$ where $\partial^\mu$ is the partial derivative (which I understand to be a vector) and $A^\mu$ is the electromagnetic vector potential (which I understand to be a vector field, the photon field upon quantisation).
From differential geometry, I understand a $2$-form to be an antisymmetric $(0,2)$-tensor field, namely an antisymmetric multilinear map $$\omega:\Gamma(TM)\times\Gamma(TM)\longrightarrow C^{\infty}(M)$$
where $\Gamma(TM)$ is the set of vector fields on $M$.
Here's my problem: I have seen in many places that "The electromagnetic field strength tensor is a $2$-form". However, I am struggling to see how the definition of the field strength tensor that I know, and the definition of a $2$-form that I know, fit together.
A $2$-form takes as its arguments two vector fields, not two vectors. $A^\mu$ is a vector field, but as far as I understand, $\partial^\nu$ is a vector (an object that takes a function and differentiates it), not a vector field. As such, it feels wrong to write the field strength tensor as something like \begin{align} F:\Gamma(TM)\times\Gamma(TM)&\longrightarrow C^{\infty}(M)\\ (\partial^\mu,A^\mu)&\mapsto F(\partial^\mu,A^\mu). \end{align} The above is obviously sloppy but I suppose what I'm saying is that it isn't clear to me that the above definition of the field strength in component form coincides in any way with the definition of a $2$-form.
I feel uneasy about my understanding of these objects in differential-geometric terms: If a derivative is a vector in differential geometry, and a tangent vector takes as its argument a function $f\in C^\infty(M)$, how/why are the derivatives in the above definition of the field strength tensor acting on the vector field $A^\mu$?
the first question
The thing is that it is a little bit sloppy to say that $F^{\mu \nu}$ is a tensor. It is a coordinate function of a tensor. The tensor, or, in this case, a 2-form, is
$$ F =\frac{1}{2} F_{\mu \nu} dx^\mu \wedge dx^\nu. $$
(Einstein summation convention is used.)
Here $dx^\mu \wedge dx^\nu = \frac{1}{2} (dx^\mu \otimes dx^\nu - dx^\nu \otimes dx^\mu)$ is antisymmetric product or exterior product. The main property of this is that it is antisymmetric, so you can neglect explicit expression via tensor products.
F acts on a pare of vector fields $(v,w)$ as
$$ F(v,w)=F_{\mu \nu} (dx^\mu v) (dx^\nu w) = F_{\mu \nu} v^\mu w^\nu. $$
Thus the answer for the first question in some coordinate system is
$$ (v,w) \mapsto F_{\mu \nu} v^\mu w^\nu. $$
the second question
I hope to clarify this place, but I am not sure if it will be satisfying. Derivatives are indeed tangent vectors, but $A^\mu$ is not a vector field, just as $F^{\mu\nu}$ is not a tensor, it is again a coordinate function of a vector field. And in this sense, it is not a problem that some vector acts on a function, though it has label $\mu$.
However, to understand this place better you should think about $A$ as about 1-form rather than vector field $A_\mu dx^\mu$. For forms (of any order) you can define exterior derivative $d$. It maps $p$-forms to $(p+1)$-forms. I'll write here how it acts in some coordinate system. Suppose you have a $p$-form $w=w_{\mu_1 \ldots \mu_p} dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_p}$, then:
$$ dw = \partial_{\mu_0} w_{\mu_1 \ldots \mu_p} dx^{\mu_0} \wedge dx^{\mu_1} \wedge \ldots dx^{\mu_p} $$
Having this, you can easily see that if $A$ is a $1$-form $A=A_\mu dx^\mu$, then
$$ d A = \partial_\nu A_\mu dx^\nu \wedge dx^\mu =\frac{1}{2} (\partial_\nu A_\mu - \partial_\mu A_\nu) dx^\nu \wedge dx^\mu=F.$$
If you are interested in details, I would recommend you to read "Gauge Fields, Knots And Gravity" by John Baez and Javier P. Muniain. It has a good elementary introduction about manifolds, vector fields, forms and all of this. Though it is not completely rigorous, it is more than enough to understand the language.