EDIT: I'm leaving this up as background reading to @drake's answer. (The point of the following is that the path integral does indeed give the correct time ordering, so it is producing the correct $\theta$-function weighted, time-ordered sums, which must be accounted for when differentiating its output.)
The two formalisms are equivalent; if they don't give the same result, something is wrong in the calculation. To see this you have to understand a subtlety which is not usually well-explained in textbooks, namely that the path integral is not defined merely by taking the limit of a bunch of integrals of the form $\int_{\mbox{lattice fields}} e^{iS(\phi)} d\phi$.
The problem is that these finite-dimensional integrals are not absolutely convergent, because $|e^{iS(\phi)}| = 1$. To define even the lattice path integral in Minkowski signature, you have to specify some additional information, to say exactly what is meant by the integral.
In QFT, the additional information you want is that the path integral should be calculating the kernel of the time evolution operator $e^{iH\delta t}$, which is an analytic function of $\delta t$. This fact is usually expressed by saying that the Minkowski signature path integral is the analytic continuation of a Euclidean signature path integral: The Euclidean $n$-point functions $E(y_1,...,y_n)$ defined by
$E(y_1,...,y_n) = \int \phi(y_1)...\phi(y_n) e^{-S_E(\phi)} d\phi$
are analytic functions of the Euclidean points $y_i \in \mathbb{R}^d$. This function $E$ can be continued to a function $A(z_1,...,z_n)$ of $n$ complex variables $z_i \in \mathbb{C}^d$. This analytic function $A$ does not extend to the entire plane; it has singularities, and several different branches. Each branch corresponds to a different choice of time-ordering. One branch is the correct choice, another choice is the 'wrong sign' time-ordering. Other choices have wrong signs on only some subsets of the points. If you restrict $A$ to the set $B$ of boundary points of the correct branch, you'll get the Minkowski-signature $n$-point functions $A|_B = M$, where $M(x_1,...,x_n) = \langle \hat{\phi}(x_1)...\hat{\phi}(x_n)\rangle_{op}$ and the $x_i$ are points in Minkowski space.
In perturbation theory, most of this detail is hidden, and the only thing you need to remember is that the $+i\epsilon$ prescription selects out the correct time-ordering.
Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say
I don't understand how rhyme these two different pictures.
Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I encourage you to try!
- For each admissible classical field configuration $\varphi:\mathbb R^3\to \mathbb R$, let $|\varphi,t\rangle$ denote a field configuration eigenstate at time $t$. Namely,
\begin{align}
\hat \phi(t,\mathbf x)|\varphi,t\rangle = \varphi( \mathbf x)|\varphi,t\rangle.
\end{align}
Take special note of the fact that $\hat\phi$ and $\varphi$ are different. The former is a operator valued distribution defined on spacetime, while the latter is a classical field configuration defined on space only.
Show that given admissible classical field configurations $\varphi_a,\varphi_b:\mathbb R^3\to \mathbb R$, there is a simple functional integral expression for the time-ordered expectation value from $|\varphi_a, t_a\rangle$ to $|\varphi_b, t_b\rangle$ of the product of a finite sequence of field operators:
\begin{align}
\langle \varphi_b, t_b| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]&|\varphi_a, t_a\rangle \\
&= \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi]} \tag{$\star$}
\end{align}
where we have defined
\begin{align}
S_{t_a,t_b}[\phi] = \int_{t_a}^{t_b}dt\int d^3\mathbf x \,\mathscr L_\phi(t)
\end{align}
and $\mathscr L_\phi$ is the Lagrangian density of the theory.
Show that the expectation value on the left hand side of $(\star)$ can be used to compute a corresponding vacuum expectation value (vev);
\begin{align}
\lim_{t\to(1-i\epsilon)\infty} \frac{\langle \varphi_b, t| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|\varphi_a, -t\rangle}{\langle \varphi_b, t |\varphi_a, -t\rangle}
&= \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle
\end{align}
where $\epsilon$ is a "positive infinitesimal" (namely you take the $\epsilon\to 0$ limit at the end). This is often called the $i\epsilon$ prescription; notice it's basically a clever trick for projecting out the ground state from a general expectation value.
Notice that the functional integration on the right hand side of $(\star)$ can be written as
\begin{align}
\int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi] }=\left(\frac{1}{i}\right)^n \frac{\delta^n Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0}
\end{align}
where we have defined
\begin{align}
Z_{t_a, \varphi_a, t_b, \varphi_b}[J] = \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,e^{iS_{t_a,t_b}[\phi]+ i\int_{t_a}^{t_b}\int d^3\mathbf x\,J(x)\phi(x)}
\end{align}
Combine steps 2-4 to show that if we define
\begin{align}
W[J] = \lim_{t\to(1-i\epsilon)\infty}\frac{Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{Z_{t_a, \varphi_a, t_b, \varphi_b}[0]},
\end{align}
then we obtain our desired expression which gives vacuum expectation values in terms of path integrals:
\begin{align}
\langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle
&= \left(\frac{1}{i}\right)^n \frac{\delta^n W[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0}
\end{align}
Notice that
\begin{align}
G^{(n)}(x_1, \dots, x_n) = \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle
\end{align}
is just a suggestive definition which makes us think of Green's functions. It's suggestive because, for example, $G^{(2)}(x_1, x_2)$, the so called "two-point function," is the Green's function for the corresponding classical field theory.
Second, you ask
Do these Feynman diagram for the two different approaches somehow represent the same scattering amplitude?
The LSZ reduction formula is the answer to the question of how vevs, or equivalently Green's functions, are related to the $S$-matrix and scattering amplitudes, and above we have argued how the canonical formalism (which is formulated in terms of vevs) is related to the functional integral formalism, so we have found how the functional integral formalism allows us to compute the $S$-matrix. In practice, it's true, that you don't see people explicitly using the LSZ reduction formula, but that's because although it conceptually underlies the connection between Green's functions and the $S$-matrix, in practice people have already used LSZ to justify codified rules, namely Feynman rules, that allow one to go directly from Feynman diagrams (which simply represent terms in the perturbative expansions of Feynman integrals) to scattering amplitudes.
Best Answer
Weinberg, QFT 2, in Section 16.1 in a footnote 2 refers to Coleman, Aspects of Symmetry, p. 135-6, which features the $\hbar$/loop expansion. See also Refs. 3 & 4 for a similar idea. In this answer we provide a non-inductive argument along these lines. A nice feature of this argument is that we do not have to deal explicitly with pesky combinatorics and symmetry factors of individual Feynman diagrams. This is already hardwired into the formalism.
A) Let us first recall some basic facts from field theory. The classical (=$\hbar$-independent) action $$S[\phi]~\equiv~\underbrace{\frac{1}{2}\phi^k (S_2)_{k\ell}\phi^{\ell}}_{\text{quadratic part}} + \underbrace{S_{\neq 2}[\phi]}_{\text{the rest}}, \tag{A1}$$ is the generating functional for bare vertices (and inverse bare propagator $(S_2)_{k\ell}$).
The partition function/path integral is $$\begin{align} Z[J] ~:=~&\int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \underbrace{\left(S[\phi]+J_k \phi^k\right)}_{=:~S_J[\phi]}\right\}\tag{A2} \cr ~\stackrel{\begin{array}{c}\text{Gauss.}\cr\text{int.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2}\cr &\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}\cr &\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}\tag{A3}\cr ~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi[J]]}{\delta \phi^m \delta \phi^n}\right)^{-1/2}\cr &\exp\left\{ \frac{i}{\hbar}\left(S[\phi[J]]+J_k \phi^k[J]\right)\right\}\cr &\left(1+ {\cal O}(\hbar)\right) \tag{A4}\end{align} $$ in the stationary phase/WKB approximation $\hbar\to 0$. In eq. (A4) $$ J_k~\approx~-\frac{\delta S[\phi]}{\delta \phi^k} \qquad \Leftrightarrow \qquad \phi^k~\approx~\phi^k[J] \tag{A5}$$ are the Euler-Lagrange (EL) equations for the quantum field $\phi^k$.
Notice in the diagram expansion (A3) how a bare vertex comes with $\hbar$-weight $=-1$; an internal bare propagator $(S_2^{-1})^{k\ell}$ comes with $\hbar$-weight $=+1$; and an external leg comes with $\hbar$-weight $=0$.
The linked cluster theorem states that the generating functional for connected diagrams is $$ W_c[J]~=~\frac{\hbar}{i}\ln Z[J], \tag{A6}$$ cf. e.g. this Phys.SE post. Note that the connected vacuum bubbles $W_c[J\!=\!0]=\frac{\hbar}{i}\ln Z[J\!=\!0]$ by definition is correlated to the normalization of the path integral, and hence is not physically relevant. (We allow the possibility that it is non-zero to be as general as possible.)
Next recall the $\hbar$/loop-expansion $$ L~=~I-V+1, \tag{A7} $$ cf. my Phys.SE answer here. The $\hbar$/loop-expansion together with eqs. (A4) & (A6) imply that the generating functional $$ W_{c}^{\rm tree}[J]~\stackrel{(A4)+(A6)}{=}~S[\phi] + J_i \phi^i \tag{A8}$$ for connected tree diagrams is the Legendre transformation of the classical action. Note that the EL eqs. (A5) are compatible with this.
Eqs. (A3) and (A6) yield $$\begin{align} &W^{\rm tree}_c[J]~\stackrel{(A3)+(A6)}{=}\cr \lim_{\hbar\to 0} \frac{\hbar}{i}& \ln\left( \exp\left\{ \frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\} \exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\} \right). \end{align}\tag{A9}$$ Notice how eq. (A9) only refers to objects in eqs. (A1) & (A8), and hence can be viewed as a consequence of them alone.
Eq. (A9) realizes the fact that given an arbitrary finite set of external source insertions, then (a sum of all possible) connected tree diagrams is (a sum of all possible) trees of bare propagators $(S_2^{-1})^{k\ell}$ and bare vertices.
Note that the one-loop square root factors in eqs. (A3) & (A4) do not affect the zero-loop/tree formula (A9) & (A8), respectively.
$\downarrow$ Table 1: Structural similarity between Sections A & B. $$ \begin{array}{ccc} A &\leftrightarrow & B \cr \phi^k&\leftrightarrow & \phi_{\rm cl}^k \cr S[\phi]&\leftrightarrow &\Gamma[\phi_{\rm cl}]\cr \hbar&\leftrightarrow &\hbar^{\prime} \cr Z[J]&\leftrightarrow &Z_{\Gamma}[J]\cr W^{\rm tree}_c[J]&\leftrightarrow &W_c[J] \end{array}$$
B) Finally let us address OP's question. Consider the effective/proper action $$ \Gamma[\phi_{\rm cl}]~\equiv~\underbrace{\frac{1}{2}\phi_{\rm cl}^k (\Gamma_2)_{k\ell}\phi_{\rm cl}^{\ell}}_{\text{quadratic part}} + \underbrace{\Gamma_{\neq 2}[\phi_{\rm cl}]}_{\text{the rest}}.\tag{B1}$$
Unlike the classical action (A1), the effective action (B1) depends (implicitly) on Planck's reduced constant $\hbar$. We would like to make a loop-expansion wrt. a new parameter $\hbar^{\prime}$.
To this end, define a partition function/path integral $$\begin{align} Z_{\Gamma}[J] ~:=~&\int \! {\cal D}\frac{\phi_{\rm cl}}{\sqrt{\hbar^{\prime}}}~\exp\left\{ \frac{i}{\hbar^{\prime}} \underbrace{\left(\Gamma[\phi_{\rm cl}]+J_k \phi_{\rm cl}^k\right)}_{=:~\Gamma_J[\phi_{\rm cl}]}\right\} \tag{B2}\cr ~\stackrel{\begin{array}{c}\text{Gauss.}\cr\text{int.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i} (\Gamma_2)_{mn}\right)^{-1/2}\cr &\exp\left\{ \frac{i}{\hbar^{\prime}} \Gamma_{\neq 2}\left[ \frac{\hbar^{\prime}}{i} \frac{\delta}{\delta J}\right] \right\}\cr &\exp\left\{- \frac{i}{2\hbar^{\prime}} J_k (\Gamma_2^{-1})^{k\ell} J_{\ell} \right\}\tag{B3}\cr ~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}& {\rm Det}\left(\frac{1}{i}\frac{\delta^2\Gamma[\phi_{\rm cl}[J]]}{\delta \phi_{\rm cl}^m \delta \phi_{\rm cl}^n}\right)^{-1/2}\cr &\exp\left\{ \frac{i}{\hbar^{\prime}}\left(\Gamma[\phi_{\rm cl}[J]]+J_k \phi_{\rm cl}^k[J]\right)\right\}\cr &\left(1+ {\cal O}(\hbar^{\prime})\right) \tag{B4}\end{align} $$ in the stationary phase/WKB approximation $\hbar^{\prime}\to 0$. Also the EL eqs. for the effective action $\Gamma_J[\phi_{\rm cl}]$ for the classical field $\phi_{\rm cl}^k$ read $$ J_k~\approx~-\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} \qquad \Leftrightarrow \qquad \phi_{\rm cl}^k~\approx~\phi_{\rm cl}^k[J]. \tag{B5}$$
Recall that the effective action (B1) is by definition the Legendre transformation of the generating functional $$ W_{c}[J]~\equiv~\Gamma[\phi_{\rm cl}] + J_k \phi_{\rm cl}^k \tag{B8}$$ for connected diagrams. Note that the EL eqs. (B5) are compatible with this.
Due to the structural similarity between two the Legendre transformations (A8) & (B8), cf. Table 1, we obtain an analogue to eq. (A9): $$\begin{align}&W_c[J]~\stackrel{(B3)+(B4)+(B8)}{=}\cr \lim_{\hbar^{\prime}\to 0} \frac{\hbar^{\prime}}{i}& \ln\left( \exp\left\{ \frac{i}{\hbar^{\prime}} \Gamma_{\neq 2}\left[ \frac{\hbar^{\prime}}{i} \frac{\delta}{\delta J}\right] \right\} \exp\left\{- \frac{i}{2\hbar^{\prime}} J_k (\Gamma_2^{-1})^{k\ell} J_{\ell} \right\} \right) .\end{align}\tag{B9}$$ In retrospect, eq. (B9) can be viewed as a functorial consequence of eqs. (B1) & (B8) alone.
On the other hand, given an arbitrary finite set of external source insertions, then (a sum of all possible) connected diagrams is (a sum of all possible) trees of full propagators$^{\dagger}$ $(\Gamma_2^{-1})^{k\ell}$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 5.
Together with eq. (B9), we conclude that the effective action $\Gamma[\phi_{\rm cl}]$ is the generating functional for (amputated) 1PI vertices (and inverse full propagator $(\Gamma_2)_{k\ell}$). $\Box$
References:
S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 16.1.
S. Coleman, Aspects of Symmetry, 1985; p. 135-6.
M. Srednicki, QFT, 2007; Chapter 21. A prepublication draft PDF file is available here.
D. Skinner, QFT in 0D, p. 32. (Hat tip: The Last Knight of Silk Road.)
P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12. (Hat tip: Abdelmalek Abdesselam.)
R. Kleiss, Pictures, Paths, Particles, Processes, Feynman Diagrams and All That and the Standard Model, lecture notes, 2013; section 1.5.2.
--
$^{\dagger}$ Fine print:
Assume that the generator $W_c[J]$ of connected diagrams has no terms linear in $J$, so that the effective action $\Gamma[\phi_{\rm cl}]$ has no terms linear in $\phi_{\rm cl}$, and so that $(\Gamma_2^{-1})^{k\ell}=-(W_{c,2})^{k\ell}$ is the full connected propagator, cf. my Phys.SE answer here.
Here the notion of the one-particle irreducible (1PI) vertices are defined wrt. to full propagators $(W_{c,2})^{k\ell}$, which is equivalent to the notion of 1PI vertices wrt. to bare propagators $(S_2^{-1})^{k\ell}$, cf. e.g. this Phys.SE post.