“Conserved” doesn’t mean “never changes”. It means “this stuff is real, and the only way you have less or more is if some is taken away or added”. You can then follow that additions or subtractions.
Since your cold bottle has less energy, the conservation law says that energy has not disappeared, it’s just gone somewhere. You can find it. You can figure out how it got there.
I have learnt that $U$ includes all forms of energy, including
translational/vibrational/rotational KE and PE.
Yes, but it is important to realize that the KE and PE components of the internal energy $U$ is the KE and PE associated with the motion and position of the molecules of the system, i.e., KE and PE at the microscopic level. The motion and positions you cannot observe directly. That is why they call it "internal" energy.
It is not the mechanical KE and PE of the system as a whole due to its velocity or position relative to an external frame of reference. I like to call this the "external" KE and PE of the system because they are associated with an external (to the system) frame of reference. These are the $\Delta KE$ and $\Delta PE$ in the general equation of the first law given in @Chet Miller answer.
Additionally, temperature is determined only by translational KE.
This statement can be misleading. Although the kinetic temperature doesn't take into account vibrational and rotational forms of molecular kinetic energy, the specific heats do take them into account, and the specific heats determine the temperature change for a given amount of heat transfer in the absence of work.
However, exercises in my textbook make it out to seem that an increase
in $U$ (whether via work or heat energy) always results in an increase
in temperature as well.
I'm not sure what those exercises are, but an increase in internal energy does not always result in an increase in temperature. For example, adding heat to cause a phase change from a solid to a liquid (e.g, melting of ice) increases internal energy with no change in temperature. That's because the heat addition only increases the internal PE, not KE of the water molecules. On the other hand, heating an ideal gas results in only an increase in translational KE.
How can work done on an object (e.g. accelerating it with a constant
force over some distance in a vacuum) increase the random
translational KE of its atoms?
Acceleration of the system as a whole increases the KE of the system as a whole, its external mechanical KE. The general form of the first law, given to you in @Chet Miller answer, tell us that the total change in energy of the object is the sum of the change in its microscopic internal PE and KE, which constitutes $\Delta U$, plus the change in its macroscopic mechanical PE and KE, the $\Delta PE$ and $\Delta KE$ in the general equation of the first law, like the thrown ball example above.
(Additional question). However, in my understanding $U$ includes both
PE and KE, so if PE was converted into KE there should be no change in
$U$.
If the conversion is between the internal (molecular) PE and internal (molecular) KE then yes, there should be no change in internal energy $U$. But it's not clear (at least to me) if they are treating the 5J of elastic potential energy as external mechanical PE of $\frac{1}{2}kx^2$ (like a spring) or internal intermolecular potential energy.
Hope this helps.
Best Answer
What you describe is a general law of conservation of energy; it assumes the rest of the universe (surroundings) can be ascribed energy and that the sum is conserved. The problem with this law is that we have no means of controlling the surroundings by definition, so it is (rather well-working) leap of faith.
The First law of thermodynamics is a little bit more restricted and more experimentally grounded. It concerns itself just with the system $S$, the surroundings are left unaccounted for.
It has more formulations which are more or less are equivalent. One of them is
Effect of heat supplied to the system on its state is equivalent to effect of certain equivalent amount of work supplied to it.$^{*}$ If both heat and work are measured in the same units (commonly Joules), a quantity characterizing the equilibrium state $X$, called internal energy, can be defined for all $X$. After the system is supplied heat $\Delta Q$ and work $\Delta W$, internal energy changes by
$$ \Delta U = \Delta Q + \Delta W. $$
(end of the law).
The change of $X$ and $U$ does not necessarily determine values of $\Delta Q$, $\Delta W$; they depend on the way the process is executed. The first law only says whatever the process (reversible, irreversible), change in $U$ is given by sum of heat and work supplied.
$^*$ The work is to be done irreversibly in such a way as to mimic addition of heat, i.e. by stirring the fluid in the system. No amount of reversible work would make the system end up in the same state that addition of heat does (since addition of heat does not conserve system's entropy but reversible work does).