Didn't we mess up with the temperature? The following passage has been extracted from the book "The Feynman Lectures on Physics-Vol l":   

The mean kinetic energy is a property only of the "temperature." Being a property of the "temperature," and not of the gas, we can use it as a definition of the temperature. The mean kinetic energy of a molecule is thus some function of the temperature. But who is tell us what scale to use to use for the temperature? We may arbitrarily define the scale of the temperature so that the mean energy is linearly proportional to the temperature. The best way to do it would be to call the mean energy itself "the temperature." That would be the simplest possible function. Unfortunately, the scale of temperature has been chosen differently, so instead of calling it temperature directly we use a constant conversion factor between the energy of a molecule and a degree of absolute temperature called a degree kelvin.  
The constant of proportionality is $k=1.38\times10^{-23}$ joule for every degree. So if T is a absolute temperature, our definition says that the mean kinetic energy is $(3/2) kt$ (The $3/2$ is put in as a matter of convenience, so as to get rid of it somewhere else.)  


From the above passage, at absolute zero, by definition, mean kinetic energy of a molecule should be zero-"completely frozen." There is a giant principle which stands against the view of atoms getting completely frozen; the following passage from the same book introduces the principle:  

As we decrease the temperature, the vibration decreases
  and decreases until, at absolute zero, there is a minimum amount of vibration
  that the atoms can have, but not zero.....  
Remember that when a crystal is cooled to absolute zero, we said that the atoms do not stop moving, they still
  jiggle. Why? If they stopped moving, we would know where they were and that
  they had zero motion, and that is against the uncertainty principle. We cannot
  know where they are and how fast they are moving, so they must be continually
  wiggling in there!  


Aren't the above two passages in contradiction with each other? Didn't we mess up with temperature?
 A: Your question, I think, is best answered by separately addressing its many directly and indirectly implied questions.  You are mixing up a lot of different notions together in your question as you've phrased it.  So to clarify, I thought I would pull all of the pieces apart.  To start:
Do atoms have nonzero kinetic energy at absolute zero?
Yes, they still posses the kinetic energy contained in their ground state.  This is colloquially known as zero-point energy.  Quantum mechanics tells us that the lowest energy state for any system has a nonzero wavefunction, and so a nonzero kinetic energy.
 
What is thermodynamics?
A completely axiomatic formulation of the theory of heat and temperature and generalized work.
For more see wikipedia.  Or for a very readable, while still short textbook, in the public domain (outside the US) see Thermodynamics by Enrico Fermi. [google books]
 
How is thermodynamic temperature defined?
Through the second law and its logical implications.
The second law has many equivalent formulations, some include Carnot's

The efficiency of a quasi-static or reversible Carnot cycle depends only on the temperatures of the two heat reservoirs, and is the same, whatever the working substance. A Carnot engine operated in this way is the most efficient possible heat engine using those two temperatures

Clausius'

Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

and Kelvin's

It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.

These all lead to the equivalent mathematical definitions:
For an ideal gas:
$$ PV = Nk_B T$$
The efficiency of a reversible heat engine operating between two thermal reservoirs is
$$ \eta = 1 - \frac{T_c}{T_h} $$
Or by the extended functional form of the first law:
$$ T = \left( \frac{ \partial U }{ \partial S } \right)_{V,N} $$
For more info, see wikipedia.
 
How is the absolute zero of thermodynamic temperature defined?
The classical theory of thermodynamics has as its third law:

The entropy of a perfect crystal, at absolute zero kelvin, is exactly equal to zero.

equivalently* it can be stated in the Nerst-Simon form:

The entropy change associated with any condensed system undergoing a reversible isothermal process approaches zero as temperature approaches 0 K, where condensed system refers to liquids and solids.

As an equation, this reads:
$$ \lim_{T \to 0} (S(T) - S(T=0)) = 0 $$
For more info, see wikipedia
 
What is statistical thermodynamics?
The kinetic and probabilistic theory of thermodynamic phenomenon.
For more information see wikipedia. For a nice readable textbook introduction, see Statistical Thermodynamics by Erwin Schrödinger [google books]
 
How is temperature defined in statistical mechanics?
Indirectly through Boltzman's definition of entropy
$$ S = k_B \log \Omega $$
and the postulate that all states are equally likely, which gives rise to the fact that states with different energies have a Boltzmann factor ratio in their populations
$$ e^{-\beta \Delta E} $$
Enforcing that statistical thermodynamics agrees with classical thermodynamics results in the identity
$$ \beta = \frac{1}{k_B T} $$
this makes the two theories equivalent in their predictions.
 
Is thermodynamic temperature equivalent to statistical mechanical temperature?
Yes. See previous questions, Thermodynamics by Enrico Fermi [google books]  or Statistical Thermodynamics by Erwin Schrödinger [google books]
 
How is absolute zero defined in statisical mechanics?
As a limiting consideration.  If the fractional populations of different states goes as a boltzmann factor $e^{-\Delta E/T}$ in the limit that $T \to 0$ all but the possibly $n$ degenerate, lowest energy states will have zero occupation, at which point the entropy becomes a constant,
$$ \lim_{T \to 0} S(T) = k_B \log n = \text{ const.}$$
independent of any parameters.  For most practical problems, this constant is so small compared to the entropies ordinarily involved that it is common to treat it as zero.  Quoting Schrödinger's book:

$$ S = k \sum_i e^{-\epsilon_i/kT} + \frac{U}{T} + \text{ const.} \qquad (2.19) $$
...
It is well known that to adopt $S=0$ at $T=0$ for every system is the conventional and most convenient way of pronouncing Nernst's famous heat theorem, sometimes called the Third Law.  Have we then, by establishing (2.19)  and by the subsequent considerations of this section, given the heat theorem a quantum statistical foundation?  At first sight it seems not, since our putting $\text{const.} =0$ was an entirely arbitrary step.
Yet we have.  For in point of fact--and contrary to what is often maintained--the numerical value of that const. is irrelevant, even meaningless.  The relevant fact is that it is a constant, in other words, that the part of the entropy which does not vanish at $T=0$ is independent of the 'parameters'. This fact entails the heat theory statistically (as we shall immediately explain) in every particular case, and so in general, provided always that we exploit the idea of 'changing parameters' in the most general determination of which it is capable.

From Statistical Thermodynamics by Erwin Schrödinger. Chapter III, pg. 16
 
How is temperature defined?
Temperature, as it is used in science is defined by the International System of Units (SI), which since 1976 has had the definition:

The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.

See the official brochure from the Bureau international des poids et mesures
This is an idealized abstract definition, to be taken in light of known physics.
 
Why is Absolute Zero -273.15 °C?
It is defined that way.  See previous and following question.
 
How do we realize this definition of temperature?
We cannot realize $T=0$, our measurements of temperature can instead only be calibrated against its extrapolated position on the temperature scale by doing some kind of experiment.  The actual definition given is an abstract one.
Recognizing that the definition of temperature is abstract, the Bureau international des poids et mesures [bipm.org] maintains curated set of mise en pratique (a list of suggested experimental realizations) for each of the defined units.  The current official mise en pratique [pdf] in its officially unofficial English translation (the french are very protective of their language) states in its introduction:

The unit of the fundamental physical quantity known as thermodynamic temperature, symbol T, is the
kelvin, symbol K, defined as the fraction 1/273.16 of the thermodynamic temperature of the triple point
of water [13th CGPM, Resolutions 3 and 4]. The International Committee for Weights and Measures
(CIPM) [Recommendation 2, CI-2005] recently clarified the definition of the triple point of water by
specifying the isotopic composition of the water to be that of Vienna Standard Mean Ocean Water
(V-SMOW). Triple-point-of-water cells provide a convenient realization of this definition. For
temperatures other than the triple point of water, direct measurements of thermodynamic temperature
require a primary thermometer based on a well-understood physical system whose temperature may be
derived from measurements of other quantities. In practice, primary thermometry is difficult and time
consuming and not a practical means of disseminating the kelvin. As an alternative, the International
Temperature Scale provides an internationally accepted recipe for realizing temperature in a practical
way.

The current official specification of the International Temperature Scale (ITS) was that outlines in 1990 in ITS-90 though there is a preliminary draft of the next version (the PLTS-2000).  In the ITS-90, the practical standard temperature is only defined down to 0.65 K (the PLTS-2000 contains provisions down to 0.9 mK) and is comprised of several overlapping scales:

*

*0.65 K - 5 K: helium vapor-pressure temperature according to:

$$T_{90}/K = A_0 + \sum_{i=1}^9 A_i \left[ (\ln (P/\text{Pa}) - B )/C \right]^i $$
where the coefficients are defined in the document.

*

*3.0 K - 24.5551 K : gas thermometer

A ${}^3$He or ${}^4$He gas thermometer calibrated against set values for the triple point of hydrogen (13.8033 K), neon (24.5561 K) and a temperature between 3 and 5 K determined by the previous scale.

*

*4.2 K - 24.5561 K : ${}^3$He, ${}^4$He as a thermionic gas

defined by: $$ T_{90} = a + b p + c p^2 $$ with given coefficients for values above 4.2 K, or taking into account the non-ideality of the gas, by $$ T_{90} = \frac{ a + b p + cp^2 }{ 1 + B_x(T_90) N/V} $$

*

*13.8033 K - 1234.93 K:  platinum resistance thermometer.

Where the temperature is determined by various functional forms for the ratio of the resistance of the platinum resistor at the desired temperature with respect to at the triple point of water.  All equations and constants given.
It goes on to give specialized calibration curves for the platinum resistor for 11 different temperature ranges in this range based on defined triple points for various substances such as  hydrogen, neon, oxygen, argon, water, and mercury, and defined values for the freezing points of various metals, such as silver, tin, zinc, aluminum, and indium, and the melting point of gallium.

*

*1240.93 K and up: Plank's radiation law

For high temperatures it is defined as:
$$ \frac{ L_\lambda(T_{90}) }{ L_\lambda [T_{90}(X) ] } = \frac{ \exp ( c_2 [ \lambda T_{90}(X) ]^{-1} ) - 1 }{ \exp (c_2 [ \lambda T_{90} ]^{-1} ) - 1 } $$
where $T_{90}(X)$ is a choosen defined metal freezing point, and $L_{\lambda}$ is the spectial concentrations of the radiance of a blackbody at the given wavelength.  Constants given.
As you can see, in practice, we don't actually observe $T=0$ (since it cannot be reached), it's point is instead extrapolated to as the limiting value in a broad range of different physical phenomenon
 
Why don't we measure temperature in units of energy?
Mostly historical.  Since time immemorial, humans have discussed some notion of temperature (or heat; the two ideas were confused until Carnot) long before we discovered its relation to kinetic energy through thermodynamics.  It is a completely intuitive notion which feels independent of energy, and so we discuss it as such.  Yes we now know they are related concepts, but just as since relativity we know time and space are related concepts, we still use distinct dimensions to denote the two in most speech, but in our calculations it is completely normal to use energy as a unit of temperature, or not, as the person doing the calculation desires.
EDIT: Since 20 May 2019, we do measure temperature in units of energy and the numerical value of $k_B$ is fixed, see 2019 redefinition of the SI base units.
 
How should we measure temperature?
That question is opinion based and would be closed.
But I think I can unambiguously point out the plethora of options, from the alternative traditional scales in common use like Fahrenheit, Celsius, Kelvin, or the Rankine (an absolute Fahrenheit scale used by some American engineers), there have been several physics based suggestions, from as the Feynman passage you mention, wherein they suggest measuring temperature in terms of energy by fixing $k_B = 1$, to some statistical mechanics books that suggest $\beta = \frac{1}{k_BT}$ as more natural, to proposals to use logarithmic temperatures $T' = \log  T/T_{\text{ref}}$ †
e.g. Pellicer J, Hernandez MJ, and Dolz M. "A new definition of the logarithmic temperature scale based on the triple point of water." REVISTA MEXICANA DE FISICA 43.5 (1997): 837-841.
A: If carefully interpreted and converted to mathematics, the passages are not contradicting one another.
The uncertainty principle guarantees that there is some zero-point energy that  can't be eliminated (the second passage) – it is the energy of the ground state of the physical object (atom or a macroscopic piece of a material).
On the other hand, the kinetic energy that increases with $T$ and reaches $E_k=0$ for $T=0$ is the excess energy above the energy of the ground state: if we want $E=0$ for $T=0$, the additive shift $\Delta E$ of the energy has to be chosen appropriately to guarantee this condition. The right additive shift is really the "subtraction of the ground state energy", i.e. of the minimum energy eigenvalue that the physical system may have.
If a physical system is frozen to $T=0$, it means that this physical system is "certain" (100%) to be found in the ground state i.e. the energy eigenstate with the minimum allowed value of $E$. In most contexts, this ground state is pretty much unique. It contains "some motion", by the uncertainty principle, but "the amount of motion above the minimum level allowed by the uncertainty principle" is zero.
In the first passage, Feynman really talks about the overall kinetic energy of the atom as it moves through space. This is indeed $3kT/2$ and strictly goes to zero for $T=0$. No subtraction is needed here. The minimum kinetic energy of the "overall motion of the object through space" is really $E_k=0$ and the corresponding momentum is $\vec p =0$. This doesn't violate the uncertainty principle because, indeed, $\Delta x = \infty$ or $\Delta x \to \infty$. The position of an atom frozen to $T=0$ (which can only be approached in the real world) is absolutely undetermined. The subtle issues of the zero-point energy only arise for the "internal energy" i.e. the relative motion of parts of a bound state (e.g. the motion of electrons around the nuclei). Only for the bound state, we really know that $\Delta x$ cannot be infinite. The constituents' being "bound" means that the distance between them is finite and bounded from above.
A: TL;DR:
Feynman was talking about a gas, so he was correct: molecules of a gas are largely independent, they don't have any quantum zero point kinetic energy.  Their mean (random) kinetic energy is temperature.
If he had been talking about any other particles such as atoms in a solid he would not have been correct - they do have kinetic energy even at absolute zero.
