Let $(E,\pi,M)$ be a bundle which coordinates is given by $(x^i,u^\alpha)$ and $J^1\pi$ the first jet associated with this bundle which coordinates is given by $(x^i,u^\alpha,u^\alpha_i)$
Given a vector vector field $\xi=\xi^i\frac{\partial }{\partial x^i}+\xi^\alpha\frac{\partial }{\partial u^\alpha}$ on $E$ its , its first lift is given by $$\xi^1=\xi^i\frac{\partial }{\partial x^i}+\xi^\alpha\frac{\partial }{\partial u^\alpha}+(\frac{d \xi^\alpha }{d x_i}-u^\alpha _j\frac{d \xi^j }{d x_i})\frac{\partial }{\partial u^\alpha _i}$$
and Lagrangian density on $\pi$ is a function $L \in C^\infty(J^1\pi)$. The corresponding Lagrangian is the m-form $\mathcal{L}=L\Omega \in \Lambda^{m}_0 \pi_1 $ where $\Lambda^{m}_0 \pi_1 $ is the space of m-forms on $J^1\pi$ horizontal over $M$. Now in this paper https://www.icmat.es/Thesis/CMartinezCampos.pdf on page 56 it has $$i_{\xi^1}d\mathcal{L}=\xi^1(L)d^mx-d\mathcal{L}\wedge i_{\xi^1}d^mx \tag{1}$$
where $i_{\xi^1}$ denotes the interior product and $d$ de exterior derivative.
I am not seeing why this formula is correct. For example if we have this bundle $(R^2,\pi,R)$ and he first jet $J^1\pi$ associated with this bundle which coordinates is given by $(x,y,z)$
then $\mathcal{L}$ would be of the form $$\mathcal{L}=Ldx$$ and $$\xi^1=\xi^x\frac{\partial }{\partial x}+\xi^y\frac{\partial }{\partial y}+\xi^z\frac{\partial }{\partial z}$$ and so $$i_{\xi^1}d\mathcal{L}=i_{\xi^1}d(Ldx)=i_{\xi^1}\Bigg(\frac{\partial L}{\partial y}dy\wedge dx+\frac{\partial L}{\partial z}dz\wedge dx\Bigg)=$$ $$=\frac{\partial L}{\partial y}dy(\xi^1)\wedge dx-\frac{\partial L}{\partial y}dy\wedge dx(\xi^1)+\frac{\partial L}{\partial z}dz(\xi^1)\wedge dx)-\frac{\partial L}{\partial z}dz\wedge dx(\xi^1)$$ $$=\xi^y\frac{\partial L}{\partial y}dx-\xi^x\frac{\partial L}{\partial y}dy+\xi^z\frac{\partial L}{\partial z}dx-\xi^x\frac{\partial L}{\partial z}dz\tag{2}$$
Can anyone explain me how expression $(1)$ and $(2)$ are related
This can be seen as an application of the product rule for the interior product and Cartan's magic formula: $$\begin{align*}i_{\xi^1}d\mathcal{L} & = i_{\xi^1}d(Ldx) = i_{\xi^1}(dL\wedge dx) = (i_{\xi^1}dL)\wedge dx-dL\wedge(i_{\xi^1}dx)\\ & = \big(\mathcal{L}ie_{\xi^1} L - d(i_{\xi^1}L)\big)\wedge dx-dL\wedge i_{\xi^1}dx\\ &= \xi^1(L)dx-dL\wedge i_{\xi^1}dx.\end{align*}$$ Here $\mathcal{L}ie$ denotes the Lie derivative and $i_{\xi^1}L=0$ because it would be a $-1$-form ($L$ is a $0$-form).
You calculated the l.-h.s. of $(1)$ explicitly in your formula $(2)$. You can do the same for the terms on the r.-h.s.: $$\xi^1(L)dx = \xi^x\frac{\partial L}{\partial x}dx + \xi^y \frac{\partial L}{\partial y}dx + \xi^z \frac{\partial L}{\partial z}dx,$$ $$dL\wedge i_{\xi^1}dx = \xi^x \frac{\partial L}{\partial x}dx + \xi^x\frac{\partial L}{\partial y}dy + \xi^x\frac{\partial L}{\partial z}dz.$$ As you can see, their difference matches the left-hand side, i.e. formula $(2)$.