If we look at the energy functional
$$E(u) = \int_\Omega L(t, u(t), u'(t), \dots, u^{(n)}(t)) dt, $$
the Euler-Lagrange equation is defined as
$$\frac{\partial L}{\partial u} + \sum_{k = 1}^{(n)} (-1)^k \frac{d^k}{d t^k} \left(\frac{\partial L}{\partial u^{(k)}}\right) = 0.$$
I found in some (German) textbooks the claim that this ODE is always of order 2$n$, as long as $L$ depends on $u^{(n)}$ and everything is well defined (i.e. we can take all necessary derivatives). However if we define
$$L(t, u(t), u^{(1)}(t), u^{(2)}(t)) = u^{(1)}(t)\cdot u^{(2)}(t)$$
we get $$ (-1) \frac{d}{dt}(u^{(2)}(t)) + \frac{d^2}{dt^2}(u^{(1)}(t)) = 0 $$ $$ \Leftrightarrow 0 = 0, $$ which isn't a ODE of order 4. By using $L(t, u(t), u^{(1)}(t), u^{(2)}(t)) = u^{(1)}(t)\cdot u^{(2)}(t) + (u^{(1)}(t))^2$, we get an ODE of order 2 (i.e $2 u^{(2)}(t) = 0$).
So, what is the order the Euler-Lagrange equation? I think that the order should be 2$n$ if $\forall \mathbf{x} \in \mathbb{R}^{n+2} : L(\mathbf{x}) \leq 0$ or $\forall \mathbf{x} \in \mathbb{R}^{n+2} : L(\mathbf{x}) \geq 0$ holds, but whats about the more general case?
What if we look at an energy function where $u$ takes two variables? I.e. $$E(u) = \int_\Omega L(x, y, u(x, y), u_x(x, y), u_y(x, y), u_{xx}(x,y),u_{xy}(x,y), u_{yy}(x,y) \dots) dx dy.$$ We write $u_x$ for the partial derivative with respect to $x$.
[EDIT: I'm especially interested if the order is always even or not]
It is true that a Lagrangian of order $n$ yields Euler-Lagrange (EL) equations of at most order $2n$.
OP's 2 example with less than maximal order $2n$ can readily be understood by identifying total derivative terms, which don't contribute to the EL equations.
If there is more than 1 $u$ variable, we can produce EL equations of odd order. This is e.g. what is happening in a Hamiltonian formulation $$L(q,\dot{q},p,t)=\dot{q}p- H(q,p,t),$$ where the EL equations (=Hamilton's equations) are of 1st order.
For the rest of this answer, we assume that there is only 1 $u$ variable.
Proposition 1. A sufficient and necessary condition for the maximal order $2n$ is that $$\frac{\partial^2L}{\partial u^{(n)}\partial u^{(n)}}\neq 0 .$$
Proposition 2. The order of the EL equation cannot be odd.
Sketched proof of Proposition 2. Assume $$\frac{\partial L}{\partial u^{(n)}}~\neq~ 0\quad\text{but}\quad \frac{\partial^2L}{\partial u^{(n)}\partial u^{(n)}}~=~ 0 . $$ Each term in $L$ of the form $u^{(n)} f(u^{(n-1)},u^{(n-2)},\ldots, u;x)$ can be rewritten into a form $g(u^{(n-1)},u^{(n-2)},\ldots, u;x)$ up to a total derivative term. In other words, we can replace the Lagrangian $L$ with an equivalent Lagrangian of order $n\!-\!1$. Repeat with $n$ replaced by $n\!-\!1$. $\Box$