Question: If I have a Lagrangian (made up equation in this case):

$$L = \frac{1}{2}mr^2\dot\theta + \frac{1}{4}mg\ddot\theta \, ,$$

can I immediately conclude that the total energy is constant because $\partial L / \partial t = 0$, because $L$ does not depend explicitly on $t$? Can I also conclude that the angular momentum is constant because $L$ does not depend explicitly on $\theta$, and that the linear momentum isn't because $L$ depends on $r$?

The notes seem to imply that one can just figure the constants of motion like so without any computations! Is this correct?

If not, what actual computations would I have to do to figure out the constants of motion from a Lagrangian?

Answer: 
If $q\_i$ denotes the generalised coordinates, then note that a time translation $t \to t+\epsilon$ infinitesimally corresponds to $q\_i \to q\_i + \epsilon \frac{d}{dt}q\_i$ and so $\delta q\_i = \dot{q}\_i$. The change in the Lagrangian is,

$$\delta L = \frac{\partial L}{\partial q\_i} \dot q\_i + \frac{\partial L}{\partial \dot q\_i} \ddot q\_i = \frac{d}{dt}L$$

which is the total derivative if $L$ has no explicit time dependence. By Noether's theorem, we have a conserved quantity,

$$Q = \frac{\partial L}{\partial \dot q\_i}\dot q\_i - L $$

which we can recognise as the Legendre transform of the Lagrangian, that is, the Hamiltonian and thus the energy of the system is conserved. Clearly this applies to your Lagrangian.

As you noted, your Lagrangian is also invariant under $\theta \to \theta + \alpha$ for which $\alpha \delta \theta = \alpha$, and $\delta L =0$. We thus have a conserved quantity which we can identify as the conjugate momentum, namely,

$$p\_\theta = \frac{\partial L}{\partial \dot \theta} = \frac12 mr^2.$$

Another way to see this is that the Euler-Lagrange equations read, $\frac{\partial L}{\partial \theta} = \frac{d}{dt}\frac{\partial L}{\partial \dot \theta}$ and since there is no $\theta$ dependence,

$$\frac{d}{dt}\frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} p\_\theta = 0.$$
Normally, observing cyclic coordinates in a Langrangian signals a conserved quantity. By cyclic I mean ignorable or basically not present. You should see Goldstein for this. 

Secondly Noether's theorem will allow you to identify constants of the motion. To answer your question though, yes we can simply observe conserved quantities by looking at the Langrangian of the system. You may enjoy looking for conserved quantities in some relativistic space time metrics such as that of Karl Schwarzschild. This could be a fun and educational example. There are many more of course. Good luck.