1) Offshell vs. onshell action. What may cause some confusion is that Noether's theorem in its original formulation only refers to the offshell action functional
$$\tag{1} I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t), $$
while Feynman's proof [1]$^1$ mostly is referring to the Dirichlet onshell action function
$$\tag{2} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl};t_i,t_f], $$
where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the extremal/classical path, which satisfies the equation of motion (e.o.m.)
$$\tag{3}\frac{\delta I}{\delta q}
~:=~\frac{\partial L}{\partial q}
 \frac{ d}{dt} \frac{\partial L}{\partial \dot{q}}~\approx~ 0,$$
with the Dirichlet boundary conditions
$$\tag{4} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f.$$
See also e.g. this Phys.SE answer. [Here the $\approx$ symbol means equality modulo e.o.m. The words onshell and offshell refer to whether e.o.m. are satisfied or not.]
2) Noether's theorem. Let us recall the setting of Noether's theorem. The offshell action is assumed to be invariant
$$\tag{5} I[q^{\prime};t^{\prime}_i,t^{\prime}_f]~=~ I[q;t_i,t_f] $$
under an infinitesimal global variation
$$\tag{6} t^{\prime}t~=~\delta t~=~\varepsilon X(t) \qquad \text{and}\qquad q^{\prime}(t^{\prime}) q(t)~=~ \delta q(t) ~=~ \varepsilon Y(t).$$
Here $X$ is a horizontal$^2$ generator, $Y$ is a generator, and $\varepsilon$ is an infinitesimal parameter that is independent of $t$.
Noether's theorem. The offshell symmetry (5) implies that the Noether charge
$$\tag{7} Q~:=~p Y  h X $$
is conserved in time
$$\tag{8} \frac{dQ}{dt}~\approx~0$$
onshell.
Here
$$ \tag{9} p~:=~\frac{\partial L}{\partial \dot{q}} \qquad \text{and}\qquad
h~:=~p\dot{q}L $$
are by definition the momentum and the energy function, respectively.
3) Assumptions. Let us assume$^3$:
that the Lagrangian $L(q,v,t)$ is a smooth function of its arguments $q$, $v$, and $t$.
that there exists a unique classical path $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ for each set $(q_f,t_f;q_i,t_i)$ of boundary values.
that the classical path $q_{\rm cl}$ depends smoothly on the boundary values $(q_f,t_f;q_i,t_i)$.
4) Differential ${\rm d}S$.
Lemma. The Dirichlet onshell action function $S(q_f,t_f;q_i,t_i)$ is a smooth function of its arguments $(q_f,t_f;q_i,t_i)$. The differential is
$$ \tag{10} {\rm d}S(q_f,t_f;q_i,t_i) ~=~ (p_f {\rm d}q_f  h_f {\rm d}t_f) (p_i {\rm d}q_i  h_i {\rm d}t_i), $$
or equivalently,
$$ \tag{11} \frac{\partial S}{\partial q_f}~=~p_f , \qquad \frac{\partial S}{\partial q_i}~=~p_i, $$
and
$$ \tag{12} \frac{\partial S}{\partial t_f}~=~h_f, \qquad \frac{\partial S}{\partial t_i}~=~h_i. $$
Proof of eq. (11):
^ q
 ____________________________
  q*_cl 
  
 ____________________________
 q_cl


> t
t_i t_f
Fig. 1. Two neighbouring classical paths $q_{\rm cl}$ and $q^{*}_{\rm cl}$.
Consider a vertical infinitesimal variation $\delta q$ between two neighbouring classical paths $q_{\rm cl}$ and $q^{*}_{\rm cl}=q_{\rm cl}+\delta q$, cf. Fig.1. The change in the Lagrangian is
$$\tag{13} \delta L
~=~ \frac{\partial L}{\partial q} \delta q
+ \frac{\partial L}{\partial \dot{q}} \delta \dot{q}
~\stackrel{(3)+(9)}{=}~ \frac{\delta I}{\delta q} \delta q
+ \frac{d}{dt}(p~\delta q)
~\stackrel{(3)}{\approx}~\frac{d}{dt}(p~\delta q),$$
so that
$$ \tag{14} \delta S ~\stackrel{(2)}{\approx}~\delta I
~\stackrel{(1)}{=}~
\int_{t_i}^{t_f}\! {\rm d}t ~\delta L
~\stackrel{(13)}{\approx}~[p~\delta q]_{t_i}^{t_f}
~=~p_f~\delta q_f p_i~\delta q_i. $$
This proves eq. (11).
Proof of eq. (12):
^ q

q*_f/
 /
 / 
 / 
q_f/ 
 / 
 /  
 q_cl/  
 /  
q_i/  
 /  
 /   
 /   
q*_i/   
    
> t
t*_i t_i t_f t*_f
Fig. 2. The classical path $q_{\rm cl}$.
Next consider the classical path $q_{\rm cl}$ between $(t_i,q_i)$ and $(t_f,q_f)$, cf. Fig. 2. Imagine that we infinitesimally extend both ends of the time interval $[t_i,t_f]$ to $[t^{*}_i,t^{*}_f]$, where
$$\tag{15}\delta t_i~:=~t^{*}_i  t_i \qquad\text{and}\qquad
\delta t_f~:=~t^{*}_f  t_f$$
both are infinitesimally small. This induces a change of the boundary positions (4) of the fixed classical path $q_{\rm cl}$ as follows
$$\tag{16} \delta q_i~:=~ q^{*}_i  q_i~=~\dot{q}_i ~\delta t_i
\qquad \text{and}\qquad
\delta q_f~:=~ q^{*}_f  q_f~=~\dot{q}_f ~\delta t_f,$$
which are dictated by the end point velocities. We would now like to calculate the variation
$$ S(q^{*}_f,t^{*}_f;q^{*}_i,t^{*}_i)  S(q_f,t_f;q_i,t_i)
~=~\delta S
~\stackrel{(11)}{=}~p_f \delta q_f +\frac{\partial S}{\partial t_f} \delta t_f p_i \delta q_i + \frac{\partial S}{\partial t_i}\delta t_i $$
$$\tag{17} ~\stackrel{(16)}{=}~(p_f \dot{q}_f +\frac{\partial S}{\partial t_f}) \delta t_f (p_i \dot{q}_i  \frac{\partial S}{\partial t_i})\delta t_i. $$
Since the new classical path is just an infinitesimal extension of the same old classical path, we may also estimate the variation as
$$ \tag{18} \delta S~=~S(q^{*}_f,t^{*}_f;q_f,t_f)+S(q_i,t_i;q^{*}_i,t^{*}_i)
~=~ L_f \delta t_f  L_i \delta t_i.$$
Comparing eqs. (9), (17) and (18) yields eq. (12).
Corollary. The Dirichlet onshell action along an infinitesimal path segment generated by the infinitesimal symmetry transformation (6) is proportional to the Noether charge
$$ \tag{19} S(q_i+\delta q,t_i+\delta t;q_i,t_i)~=~\varepsilon Q_i.$$
Proof of the Corollary:
$$ \tag{20} S(q_i+\delta q,t_i+\delta t;q_i,t_i)
~\stackrel{(10)}{=}~p_i\delta q h_i \delta t
~\stackrel{(6)}{=}~\varepsilon(p_i Y h_i X)
~\stackrel{(7)}{=}~\varepsilon Q_i.$$
5) Feynman's fourpoint argument. We are finally ready to discuss Feynman's fourpoint argument.
^ q

 A' B'
 ___________________________
  virtual/offshell 
  
  
 ___________________________
 A classical/onshell B


> t
Fig. 3. Feynman's four points. (Note that the two horizontal and the two vertical straight ASCII lines are in general an oversimplification of the actual paths.)
We start with the onshell action
$$\tag{21} S(A\to B)~=~I(A\to B)$$
for some classical path $q_{\rm cl}$ between two spacetime events $A$ and $B$. We then apply the infinitesimal transformation (6) to produce a virtual path $q^{\prime}$ between two infinitesimally shifted spacetime events $A^{\prime}$ and $B^{\prime}$. In turn, the virtual path $q^{\prime}$ has an offshell action
$$\tag{22} I(A^{\prime}\to B^{\prime})~=~I(A\to B)$$
equal to the original action due to the offshell symmetry (5).
Next we would like to consider the shifted path $A\to A^{\prime}\to B^{\prime}\to B$. Unfortunately, the two infinitesimal pieces $A\to A^{\prime}$ and $B^{\prime}\to B$ (which we will choose to be classical paths) may correspond to constant time. [The timeintegration in the definition (1) of the offshell action $I(A\to A^{\prime}\to B^{\prime}\to B)$ would not make sense in case of constant time.] In such cases we replace Feynman's four points with six points, i.e. we extend infinitesimally the original classical path $A\to B$ to a classical path $A^{*}\to B^{*}$, in such a way that the two new infinitesimal paths $A^{*}\to A^{\prime}$ and $B^{\prime}\to B^{*}$ (which we also will choose to be classical paths) do both not correspond to constant time.
^ q

 A' B'
 ____________________________
 / virtual/offshell \
 /   \
 /   \
 A* /_________________________________\ B*
 A classical/onshell B


> t
Fig. 4. Six points.
Since the virtual path $A^{*}\to A^{\prime}\to B^{\prime}\to B^{*}$ is an infinitesimal variation of the classical path $A^{*}\to A\to B\to B^{*}$, we conclude that the difference
$$S(A^{*}\to A^{\prime})+I(A^{\prime}\to B^{\prime})+S(B^{\prime}\to B^{*})$$
$$S(A^{*}\to A)S(A\to B)S(B\to B^{*})$$
$$\tag{23} ~=~I(A^*\to A^{\prime}\to B^{\prime}\to B^*)
S(A^*\to A\to B\to B^*)~=~{\cal O}(\varepsilon^2)$$
cannot contain contributions linear in $\varepsilon$.
We next apply the Lemma and Corollary from Section 4. The six infinitesimal classical paths mentioned so far are all described by the differential (10), which is linear and hence obeys a (co)vector addition rule. Therefore
$$\tag{24} S(A^{*}\to A^{\prime})S(A^{*}\to A) +{\cal O}(\varepsilon^2)
~\stackrel{(10)}{=}~S(A\to A^{\prime})
~\stackrel{(19)}{=}~\varepsilon Q_i,\qquad $$
$$\tag{25} S(B\to B^{*})  S(B^{\prime}\to B^{*})+{\cal O}(\varepsilon^2)
~\stackrel{(10)}{=}~S(B\to B^{\prime})
~\stackrel{(19)}{=}~\varepsilon Q_f.\qquad $$
Comparing eqs. (21)(25), we arrive at the main conclusion of Noether's theorem,
namely that the Noether charge is conserved,
$$\tag{26} Q_f~=~Q_i.$$
References:
 R.P. Feynman, The Character of Physical Law, 1965, pp. 103  105.

$^1$ For Feynman's proof, see approximately 50 minutes into this video. Noether's theorem is covered in 45:2551:27.
$^2$ Feynman uses the opposite convention for horizontal and vertical than this answer.
$^3$ Noether's theorem works with less assumptions, but to avoid mathematical technicalities, we impose assumption 1, 2 and 3. Note that it is easy to find examples that satisfies assumption 1 and 2, but where the classical path $q_{\rm cl}$ may jump discontinuously for varying boundary values $(q_f,t_f;q_i,t_i)$, so that assumption 3 is not satisfied.
This post imported from StackExchange Physics at 20150503 09:39 (UTC), posted by SEuser Qmechanic