I have a function of two variables $f(x,y)\ ,\ x,y\in \mathbb{R}$ that I allow to be discontinuous on the second variable $y$ (or both it doesn't matter) , I am allowed to compute $F(x,y)= \frac{\partial}{\partial y} f(x,y)$ as a distribution.
So far $$F: \phi\mapsto \int_{\mathbb{R}^2} \phi(\textbf{x})\frac{\partial}{\partial y} f(u,v)\,d\textbf{x}\ ,\quad \phi\in C^\infty_0(\mathbb{R}^2)$$ is well defined, where $\textbf{x}= (u,v)\in \mathbb{R}^2$ and $\phi$ is a smooth compactely supported function on $\mathbb{R}^2$
is this definition correct ?
Now I need to integrate over the first variable $x$, that is, I want to compute $$\int_{\mathbb{R}} F(x,y)\,dx \equiv \int_{\mathbb{R}} \frac{\partial}{\partial y} f(x,y)\,dx $$ Let us assume here the integral is always converging. How to compute this in practice, I am a bit confused since $F$ is a distribution and I don't quite understand integration of distributions.
What is $\int_{\mathbb{R}} F(x,y)\,dx\quad $ in term of $\quad F(\phi)= \int_{\mathbb{R}^2} \phi(\textbf{x})\frac{\partial}{\partial y} f(u,v)\,d\textbf{x}\ $ ??
I think you are confusing two different "integrations".
First, as @MarkViola comments, the action of a distribution is not literally by integration, although that it the spirit of the thing. If your $f$ is locally $L^1$ in both variables, then its partial derivative in $y$ does act on test functions $\varphi$ by ${\partial f\over\partial y}(\varphi)=-\int f\,{\partial \varphi\over \partial y}$, since the action of $f$ itself as a distribution is by integration.
The issue of integrating ${\partial f\over \partial y}$ with respect to $x$ is a different thing. Here, we'd be thinking of ${\partial f\over \partial y}$ as a distribution "in $y$" parametrized by $x$. And we could/should think of the original $f$ in the same way.
When $f$ is not toooo bad a function of $x$, we can apply notion of distribution-valued integral of a parametrized family of distributions (see either Gelfand-Pettis (weak) integrals, and/or Bochner integrals). In that case, the integral in $x$ of $f(x,y)$ or its partial with respect to $y$ would be a distribution in a single variable ("$y$"), not two variables. Assuming the mild hypotheses for existence of such a distribution-valued integral, with $\varphi$ a test function in one variable, we do have literal integrals: first, a distribution-valued one, but then two scalar-valued ones: $$ \Big(\int {\partial f\over \partial y}\,dx\Big)(\varphi) \;=\; \int {\partial f\over \partial y}(\varphi)\,dx \;=\; - \int f(x,-)\Big({\partial\varphi\over \partial y}\Big)\,dx $$ The Gelfand-Pettis (weak) property is essentially that the first equality is correct.