I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation.

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

Continuation: Perhaps it's worthwhile making this a little more explicit. The simplest example is $X=\mathbb{A}^n$. Then the (global sections of) $D_X$ is the Weyl algebra with generators $x_1,\ldots, x_n, \partial_1,\ldots, \partial_n$ with commutation relations $[x_i,\partial_j]= \delta_{ij}$. If we force these to commute, by passing to the associated graded with respect to the filtration by order of operators, we obtain the polynomial ring in $2n$ variables or in other words the coordinate ring of $T^*\mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)\subset T^*\mathbb{A}^n$ (it is independent of the filtration). Now let $M= D/\sum D\partial_i$, which corresponds to $\mathcal{O}_X$. I'll omit the details, but one can see $Ch(M) = V(\partial_i)$ (the zero section), and $\pi^*supp(M) = T^*\mathbb{A}^n$ where $\pi:T^*\mathbb{A}^n\to \mathbb{A}^n$

I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation. (It looks like YBL gives a clear picture of the holonomic case.)

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

Continuation: Perhaps it's worthwhile making this a little more explicit. The simplest example is $X=\mathbb{A}^n$. Then (the global sections of) $D_X$ is the Weyl algebra with generators $x_1,\ldots, x_n, \partial_1,\ldots, \partial_n$ with commutation relations $[x_i,\partial_j]= \delta_{ij}$. If we force these to commute, by passing to the associated graded with respect to the filtration by order of operators, we obtain the polynomial ring in $2n$ variables or in other words the coordinate ring of $T^*\mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)\subset T^*\mathbb{A}^n$ (it is independent of the filtration). Now let $M= D/\sum D\partial_i$, which corresponds to $\mathcal{O}_X$. I'll omit the details, but one can see that $Ch(M) = V(\partial_1,\ldots \partial_n)$ (the zero section), and $\pi^*supp(M) = T^*\mathbb{A}^n$ where $\pi:T^*\mathbb{A}^n\to \mathbb{A}^n$.

For a nonholonomic example, take $M= D_X$. Then $Ch(M)= \pi^{-1}supp(M) = T^*\mathbb{A}^n$.

I'm not a $D$-module person, so the following statements should be taken with grain of salt.

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation.

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

Continuation: Perhaps it's worthwhile making this a little more explicit. The simplest example is $X=\mathbb{A}^n$. Then the (global sections of) $D_X$ is the Weyl algebra with generators $x_1,\ldots, x_n, \partial_1,\ldots, \partial_n$ with commutation relations $[x_i,\partial_j]= \delta_{ij}$. If we force these to commute, by passing to the associated graded with respect to the filtration by order of operators, we obtain the polynomial ring in $2n$ variables or in other words the coordinate ring of $T^*\mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)\subset T^*\mathbb{A}^n$ (it is independent of the filtration). Now let $M= D/\sum D\partial_i$, which corresponds to $\mathcal{O}_X$. I'll omit the details, but one can see $Ch(M) = V(\partial_i)$ (the zero section), and $\pi^*supp(M) = T^*\mathbb{A}^n$ where $\pi:T^*\mathbb{A}^n\to \mathbb{A}^n$