First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$.
At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second definition obeys the first differential equation.
The differential operator $\hat L$ is what appears in the linearized equations of motion for the field, in this case $\psi(x_1,t_1)$, and it only acts on $\psi(x_1,t_1)$, not $\psi^\dagger (x_2,t_2)$.
The time-ordering operator $T$ may be written in terms of the step function
$$ T(\psi \psi^\dagger) = \psi \psi^\dagger\cdot \theta(t_1-t_2) - \psi^\dagger \psi\cdot\theta(-t_1+t_2)$$
where $\psi$ is always at $x_1,t_1$ and $\psi^\dagger$ is at $x_2,t_2$. Now, ask what happens when you act with $\hat L$ on the right hand side of the displayed equation above.
By the Leibniz rule, there are the terms with $\hat L \psi = 0$. It vanishes by the equations of motion. But there are extra terms where $\hat L$ acts on the step functions.
The operator $\hat L$ contains the term that differentiates with respect to $t_1$ multiplied by a coefficient $C$. This turns $\theta(t_1-t_2)$ to $\delta(t_1-t_2)$. The same occurs in the next term, but with the opposite sign which cancels the sign that was already there. So the extra terms are
$$ \hat L T(\psi \psi^\dagger) = C\delta(t_1-t_2) (\psi \psi^\dagger + \psi^\dagger \psi)$$
I got two terms because there were two terms. However, these two terms exactly combine to the anticommutator of $\psi$ and $\psi^\dagger$ which only needs to be evaluated for $t_1=t_2$, the equal-time anticommutator, and the result is $D\cdot \delta(x_1-x_2)$.
That's why the action of $\hat L$ on the correlator ends up being $CD\delta(t_1-t_2)\delta(x_1-x_2)$ where the constants $C,D$ are mostly just factors of $i$ etc.
For bosonic fields, $\hat L$ has the second derivative with respect to time. One of the derivatives has the fate as above, the other one turns the other $\phi$, which plays the role of $\psi^\dagger$, into $\partial_t \phi$ which is the canonical momentum, and it's the right variable that has the $\delta$-function-like commutator. Also, the intermediate sign is the opposite one but the result is the same, some $CD\cdot\delta\cdot\delta$.