This is quite simple. To better visualize Newton's law let's not use generalized coordinates, instead, let's use Cartesian ones. Euler-Lagrange equations hence stands:
$$
\frac{d}{dt}\frac{\partial L}{\partial\dot x} - \frac{\partial L}{\partial x} = 0
$$
Since $L = T - V$, it is clear to know that:
$$
\frac{\partial L}{\partial x} =
\frac{\partial T}{\partial x} - \frac{\partial V}{\partial x} =
-\frac{\partial V}{\partial x} \quad\Longrightarrow\quad
\frac{\partial L}{\partial x} = -\frac{\partial V}{\partial x} = F
$$
Now, on the other term, the potential doesn't depends on the velocity. Hence:
$$
\frac{\partial L}{\partial \dot x} =
\frac{\partial T}{\partial \dot x} - \frac{\partial V}{\partial \dot x} =
\frac{\partial T}{\partial \dot x} \quad\Longrightarrow\quad
\frac{\partial L}{\partial \dot x} =
\frac{\partial T}{\partial \dot x} = p
$$
This can be easily seen:
$$
T = \frac{1}{2}m\dot x^2 \Longrightarrow
\frac{\partial T}{\partial \dot x} = m\dot x = p
$$
Taking those results and plugging into Euler-Lagrange equations, we have:
$$
\frac{dp}{dt} - F = 0
$$
Which are, Newton's law.
Your question depends on the problem and the context. I'll try to answer your question in general below (but will also answer the specific question you had).
In classical mechanics we are usually interested in determining the time-evolution of particles, i.e. we are interested in finding functions $q(t)$ or more generally $\vec{q}(t) = \{q_1(t),\ldots,q_n(t)\}$. For this kind of problems the Lagrangian is a function of only $q$ and $\dot{q}$ and this always leads to second order ordinary differential equations in $q$. The reson we cannot have a PDE here is that we only have one coordinate (time). When you write the Euler-Lagrange equation as
$$\frac{dL(q,\dot{q})}{dq} - \frac{d}{dt}\frac{dL(q,\dot{q})}{d\dot{q}} = 0$$
you are in this class of problems so for your situation the answer is ODE.
In general this is not the case. For example, when working with fields (e.g. the electromagnetic field, quantum fields, the gravitational field) instead of particles, one usually ends up with partial differential equations. The Euler-Lagrange equation in these cases is more complicated. If the Lagrangian is a function of the field $q$ and the derivatives of the fields $\frac{dq}{dt}$, $\frac{dq}{dx}$, $\ldots$ then we have
$$\frac{dL(q,\nabla q)}{dq} - \nabla_\mu\left(\frac{dL(q,\nabla q)}{d\nabla_\mu q}\right) = 0$$
where the last term is a sum over the different coordinates $\mu = t,x,y,z, $ and so on.
"The Lagrange equation is a second order differential equation."
In general the Euler-Lagrange equation does not have to be a second order ODE (or PDE). For example if the Lagrangian depends on $q,\dot{q}$ and $\ddot{q}$ then the Euler-Lagrange equation becomes
$$\frac{dL}{dq} - \frac{d}{dt} \frac{dL}{d{\dot q}}+ \frac{d^2}{dt^2}\frac{dL}{d{\ddot q}} = 0$$
which can lead to a third order equation (and so on). However for physics we almost always work with Lagrangians that lead to second order ODEs (or second order in time for the PDEs). The reason for this is that Lagrangians that leads to higher order equations of motion are plagued by instabillities and it seems that nature hates instabillities just as much as physicists.
Best Answer
Let's suppose the units of $q$ are are $\text{u}$, which stands for "user units."
The units of $\frac{\partial L}{\partial q}$ are $\text{J}\text{u}^{-1}$ (Joules per user unit.)
The units of $\frac{\partial L}{\partial \dot q}$ are Joules per (user units per second) which is $\text{J}\text{u}^{-1}\text{s}$. Hence the units of $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)$ are $\text{J}\text{u}^{-1}$.
You could think of them as directional derivatives, but I think it is more helpful just to think of them as partial derivatives.