You start by writing down the probability to find a particle at $y$ at time $t$ when it was at $x$ at time $0$, denoted as $K(y,t;x,0)$.
You get this by solving the Schrödinger equation with the initial condition $\psi(y,0) = \delta(y-x)$. Then, $K(y,t;x,0) = \psi(y,t)$.
Thus, to solve this, we need to know the time development of the initial condition $\psi(y,0)$.
Let us start with the simple example of a free particle. This is easiest solved in momentum-representation, obtained by Fourier-transforming $\psi(y,t)$:
$$\psi(y,t) = \frac{1}{\sqrt{2\pi\hbar}} \int dp \exp(ipy/\hbar) \tilde \psi(p,t)$$
For $\tilde \psi$, the Schrödinger equation gives
$$\tilde \psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \exp \left(-\frac{i}{\hbar}
\left[\frac{p^2 t}{2m} - px\right]\right)$$
This can be inserted back into the equation for $\psi(y,t)$. The integral over $p$ can be solved exactly. The final result is
$$K_\text{free}(y,t;x,0) = \sqrt{\frac{m}{2\pi i\hbar t}} \exp\left(\frac{im(x-y)^2}{2\hbar t}\right)$$
Next step: The solution of the Schrödinger equation can generally be written as
$$|\psi, t\rangle = \exp\left(-\frac{iHt}{\hbar}\right) |\psi,0\rangle$$
with $H$ being the Hamiltonian of your system.
Writing $H = T+V$, the general formula for $K$ becomes
$$K(y,t;x,0) = \langle y \mid \exp(-\frac{i(T+V)t}{\hbar}) \mid x \rangle$$
We use the Trotter-Kato Formula (which holds under certain conditions which I won't go into detail at this point. It allows us to write
$$K(y,t;x,0) = \lim_{N\rightarrow \infty} \langle y \mid \left[ \exp(-\frac{iTt}{N\hbar})
\exp(-\frac{iVt}{N\hbar})\right]^N \mid x\rangle$$
We insert the unity operator, decomposed as $1 = \int dx | x \rangle \langle x |$ $N-1$ times, which gives us
$$K(y,t;x,0) = \int dx_1 dx_2 \dots dx_{N-1} \prod_{j=0}{N-1}
\langle x_{j+1} \mid \exp(-iTt/N\hbar) \exp(-iVt/N\hbar) \mid x_j \rangle$$
Note that $V$ as an operator acting on $|x\rangle$ gives just $V(x) |x\rangle$. And
$\langle x_{j+1} | \exp(-iTt/N\hbar) | x_j \rangle$ gives us just the contribution of a free particle, i.e. $$\sqrt{\frac{mN}{2\pi i\hbar t}} \exp\left(\frac{imN}{2\hbar t}(x_{j+1} - x_j)\right)^2$$.
If we abbreviate $\tau = t/N$, we can write:
$$K(y,t;x,0) = \lim_{N\rightarrow \infty} \int dx_1 dx_2 \dots dx_{N-1}
\left( \frac{m}{2\pi i\hbar \tau}\right)^{N/2} \times$$
$$\exp \left(\frac{i\tau}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\tau}\right)^2 - V(x_j)\right]\right)$$
The next step is to see the values $x_j$ as points of a certain path $x(t')$ evaluated at points
$t' = t_j = j\tau = jt/N$. If $\tau$ is small, we write
$$\sum_{j=0}^{N-1} \tau f(t_j) \rightarrow \int f(t') dt'$$
$$\frac{x_{j+1} - x_j}{\tau} \rightarrow \dot x(t')$$
where the dot denotes the time-derivative.
The argument of the exponential then becomes
$$\frac{i}{\hbar} \int_0^t dt' \left( \frac{m\dot x(t')^2}{2} - V(x(t'))\right)$$
You will have no trouble identifying the integrand as the Lagrangian $L = T-V$. The integral itself, therefore, is the classical action.
Thus, the formula we have for $K$ can be interpreted as the sum over all possible paths from $(x,0)$ to $(y,t)$ of the function $\exp\left(\frac{i}{\hbar} S(t,0)\right)$ of the classical action.
The interpretation of this was given in other answers: The classical path is that which minimizes the action, i.e. the action is stationary for the classical path. In your path-integral formula, this path will have a large contribution, as all paths that vary only slightly from the classical path will still have pretty much the same phase factor as the classical one, leading to constructive interference of those paths. For paths far from the classical path, the action will vary greater among the paths, so that there all possible phases occur, which will ultimately cancel each outer out.
Reference A lecture on advanced quantum mechanics given by Prof. Crispin Gardiner. Lecture notes are, unfortunately, not freely available. It was a good lecture :)
