The minimal polynomial is quite literally the smallest (in the sense of divisibility) nonzero polynomial that the matrix satisfies. That is to say, if $A$ has minimal polynomial $m(t)$ then $m(A)=0$, and if $p(t)$ is a nonzero polynomial with $p(A)=0$ then $m(t)$ divides $p(t)$.

The characteristic polynomial, on the other hand, is defined algebraically. If $A$ is an $n \times n$ matrix then its characteristic polynomial $\chi(t)$ must have degree $n$. This is not true of the minimal polynomial.

It can be proved that if $\lambda$ is an eigenvalue of $A$ then $m(\lambda)=0$. This is reasonably clear: if $\vec v \ne 0$ is a $\lambda$-eigenvector of $A$ then
$$m(\lambda) \vec v = m(A) \vec v = 0 \vec v = 0$$
and so $m(\lambda)=0$. The first equality here uses linearity and the fact that $A^n\vec v = \lambda^n \vec v$, which is an easy induction.

It can also be proved that $\chi(A)=0$. In particular that $m(t)\, |\, \chi(t)$.

So one example of when (1) occurs is when $A$ has $n$ distinct eigenvalues. If this is so then $m(t)$ has $n$ roots, so has degree $\ge n$; but it has degree $\le n$ because it divides $\chi(t)$. Thus they must be equal (since they're both monic, have the same roots and the same degree, and one divides the other).

A more complete characterisation of when (1) occurs (and when (2) occurs) can be gained by considering Jordan Normal Form; but I suspect that you've only just learnt about characteristic and minimal polynomials so I don't want to go into JNF.

Let me know if there's anything else you'd like to know; I no doubt missed some things out.