Clarification for the meaning of arc lenght and area forms 
Let $C$ be an oriented regular curve and $ds$ its arc lenght. If $\alpha:I\to C$ is a parametrization compatible with the orientation of $C$, then $$\alpha^*(ds)=\|\alpha'\|dt$$
  Let $S\subset\mathbb{R}^3$ be an oriented regular surface and $dA$ its area form. If $\psi:U\subset\mathbb{R}^2\to S$ is a parametrization compatible with the orientation of S then $$\psi^*(dA)=\|\psi_u×\psi_v\|du \wedge dv$$

I (more or less) understand forms on open subsets of $\mathbb{R}^n$ and their pullback and I just started with forms on manifolds. I don't think I understand what is the meaning of $ds$ and $dA$ in those statements.
I thought $ds$ should be the 1-form of $C$ whose $p-$component ($p\in C$) is $(\alpha'(q))^*$, where $p=\alpha(q)$. So, the pullback of $ds$ by $\alpha$ should be a 1-form on $I$ whose $q-$component acting on a vector of the tangent space of $I$, which in this case is just some $x \in \mathbb{R}$, is $ds_p=(\alpha'(q))^*$ acting on $\alpha'(q)x$, which is simply $$(\alpha'(q))^*[\alpha'(q)x]=x(\alpha'(q))^*[\alpha'(q)]=x$$
So maybe I'm missunderstanding the meaning of $ds$ (and similarly with $dA$)
How should I interpret those differential forms?
 A: I'm not sure I understand what your problem is, but

How should I interpret those differential forms?

is easily answerable.
Let $M$ be a smooth $n$-dimensional manifold and let $g$ be a Riemannian metric on $M$. If $(U;x^1,...,x^n)$ is a local chart, then the differential form whose local expression is $$ đ\mu_g=\sqrt{\det g}dx^1\wedge...\wedge dx^n $$ is a globally well-defined volume form, which we may call a Riemannian volume form.

It is not customary to treat the "initial space" of curves abstractly (eg. they are usually taken to be real intervals), however I can treat it like that if you want.
Let $$ \gamma:C\rightarrow M $$ be a curve, where $C$ is a one-dimensional manifold. We may then pull back the metric $g$ to $C$ to obtain $$ q=\gamma^*g. $$ But then $q$ is a Riemannian metric on $C$, so we can contruct the associated volume element (on $C$), which we call $đs$. You seemed to want to treat local coordinate expansions via pullbacks. That is certainly possible, but in my opinion, it just overcomplicates stuff.
Let $(U;x^1,...,x^n)$ be a chart on $M$ that contains the image of the curve (or at least the part of the image of the curve we care about), and let $(I;t)$ be a chart on the curve. The metric $g$ expands locally as $$ g=\sum_{\mu,\nu}g_{\mu\nu}dx^\mu\otimes dx^\nu, $$ and now we pull it back: $$ q=\gamma^*g=\sum_{\mu\nu}(g_{\mu\nu}\circ\gamma)d(x^\mu\circ\gamma)\otimes d(x^\nu\circ\gamma)=\sum_{\mu\nu}g_{\mu\nu}(x(t))\frac{dx^\mu}{dt}dt\otimes\frac{dx^\nu}{dt}dt \\ =\sum_{\mu\nu}g_{\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}dt\otimes dt,$$ where I have been progressively abusing notation more and more, but I hope it is understandable what I meant. We also see that the lone component of $q$ in the chart $(I;t)$ is just $g(\gamma',\gamma')$ where $\gamma'$ is the velocity vector of the curve $\gamma$.
So we have $$ q=\left\|\gamma'\right\|^2dt\otimes dt, $$ then we have $$ \sqrt{\det q}=\|\gamma'\|, $$ so the local expression of the "volume element" on $C$ is $$ đs=\|\gamma'\|dt $$

Likewise, if $\Sigma$ is a two-dimensional manifold and $\psi:\Sigma\rightarrow M$ is an embedding into $M$, then we define the internal metric on $\Sigma$ by $$ h=\psi^*g. $$ For local charts $(U;x^1,...,x^n)$ and $(V;\xi^1,\xi^2)$ in $M$ and $\Sigma$ respectively, we can calculate $$ h=\sum_{a,b=1}^2\sum_{\mu,\nu=1}^{n}g_{\mu\nu}\frac{\partial x^\mu}{\partial\xi^a}\frac{\partial x^\nu}{\partial\xi^b}d\xi^a\otimes d\xi^b\equiv\sum_{a,b}h_{ab}d\xi^a\otimes d\xi^b. $$
The area element is then the 2-form that is locally $$ dA=\sqrt{\det h}\ d\xi^1\wedge d\xi^2. $$ You see that $h_{ab}$ is essentially $$ g(\psi_a,\psi_b), $$ where $\psi_a$ is the (pushforward of the) tangent vector $\partial/\partial\xi^a$. We also know from linear algebra that the area of the paralelogram generated by two vectors is $$ \|u\times v\|=\sqrt{\|u\|^2\|v\|^2-\langle u,v\rangle^2}, $$ so we have $$ dA=\|\psi_1\times\psi_2\|d\xi^1\wedge d\xi^2. $$
