Context:
In calculus that I did before I would just calculate $\int \int \int 1 dx dy dz$ by calculating $\int 1 dx = x$ and that leaves me with $\int \int x dy dz$. In the new calculus I have to do of differential forms I have to integrate something like $\int \int 1 dx \wedge dy + 1 dy \wedge dz$. The wedge product $\wedge$ is a generalization of the vector product $\times$ I use for calculating the area element by which to multiply vector and scalar fields but there.
Further on in the articles it says the coordinate differentials are covectors, so I do not understand why or how they're used in an external product like this.
Short answer: $\iint 1 \,dx \wedge dy$ for a small patch means calculate the area of the projection of the patch onto the $x{-y}$ plane.
Medium answer: a differential form $dx$, also called a covector, is "a measure of distances in the $x$ direction". To say the point $(3,0)$ is 3 metres east of $(0,0)$, you need a direction "increasing $x$ coordinate is east" and a length scale "one unit is one metre", but you don't need to know where the zero line is on the east-west scale. The covector $dx$ represents this "coordinate-like thing where you can measure change in position in some direction, but without an actual position", i.e. without a fixed zero point.
Then the wedge product $\iint 1 \, dx\wedge dy$ says "if one unit in the positive $x$ direction is one metre east and one unit in the positive $y$ direction is one unit north, how many square metres in the east-north direction is our surface?"
(Aside: in specific examples, i.e. if you're given a surface via its coordinates in $\mathbb{R}^3$, it might be simple to just use the zero point from $\mathbb{R}^3$ and avoid such a confusing object as $dx$, but then you'd still want to prove that changing your zero point doesn't change the answer (like most properties of manifolds don't change when you change the coordinate chart). Theoretical mathematicians decided to create an object with no zero point, and worked out the rules for using them, so they can avoid checking every time to say that changing the zero point doesn't matter.)
Long answer: $dx$ is actually a covector field, i.e. it gives you a covector $dx_p$ at each point $p \in \mathbb{R}^3$. A covector is a linear operator on vectors, i.e. $dx_p(v)$ is linear for vectors $v \in T_p\mathbb{R}^3$ (think $v \in \mathbb{R}^3$ if you don't recognise $T_p\mathbb{R}^3$. Also, note that every linear operator on vectors in $\mathbb{R}^3$ is just the dot product with another vector in $\mathbb{R}^3$, but changing the coordinate system on $\mathbb{R}^3$ doesn't affect $dx(v)$ even though it would affect $v \cdot v$.) So $dx$ is given by $dx_p((v_1,v_2,v_3))=v_1$ for this particular coordinate system, but $dx$ is coordinate independent, and in another coordinate system we might have $dx_p((v_1,v_2,v_3))=2v_1$ (for example).
Given a covector field $dx$, we can measure the "distance in the $x$ direction" of a curve $\gamma:[0,1] \to \mathbb{R}^3$ with derivative $\dot{\gamma}(t)$ for $t \in [0,1]$. At each $t$, the curve is moving in the $x$ direction at speed $dx_{\gamma(t)}(\dot{\gamma}(t))$. So the total movement in the $x$ direction is $\int_0^1 dx_{\gamma(t)}(\dot{\gamma}(t))\,dt$. But we can also measure the total change (positive or negative) in $x$ position along the curve. Let $I:=\{\gamma(t):t \in [0,1]\}$ denote the image of $\gamma$. Then we can define $\int_I 1\,dx:=\int_0^1 |dx_{\gamma(t)}(\dot{\gamma}(t))|\,dt$, i.e. we add up the total positive or negative movement in the $x$ direction for each infinitesimal part of the curve. This is sometimes written $\int_I |dx|$ instead of $\int_I 1 \, dx$. Here, for subsets $A \subset \mathbb{R}^3$ with a suitable parametrisation, $A \mapsto \int_A 1\,dx$ is a measure (which for your purposes just means "always nonnegative", but might help you looking up definitions).
The wedge product $dx\wedge dy$ is a 2-form field, i.e. at each point $p \in \mathbb{R}^3$, it gives a bilinear function $dx_p \wedge dy_p:T_p\mathbb{R}^3 \times T_p \mathbb{R}^3 \to \mathbb{R}$. In the usual coordinate system on $\mathbb{R}^3$, we have $dx_p \wedge dy_p((v_1,v_2,v_3),(w_1,w_2,w_3))=v_1 w_2$. Given a parametrisation $\mathbf{p}:[0,1] \times [0,1]\to \mathbb{R}^3$ of a surface, we can work out $\int_0^1 \int_0^1 dx_{\mathbf{p}(s,t)} \wedge dy_{\mathbf{p}(s,t)}(\frac{\partial \mathbf{p}(s,t)}{\partial s},\frac{\partial \mathbf{p}(s,t)}{\partial t})\,dt\,ds$. We can also work out the measure of the surface, $\iint_S 1\,dx \wedge dy:=\int_0^1 \int_0^1 |dx_{\mathbf{p}(s,t)} \wedge dy_{\mathbf{p}(s,t)}(\frac{\partial \mathbf{p}(s,t)}{\partial s},\frac{\partial \mathbf{p}(s,t)}{\partial t})|\,dt\,ds$, i.e. add up the area in the $x{-y}$ plane of each patch.