Assume that Problem $A$ is polynomial-time reducible to problem $B$.
Claim 1: If problem $A$ is NP-hard then problem $B$ is NP-hard.
Claim 2: If problem $B$ is NP-hard then problem $A$ is NP-hard.
Claim 1 is obviously true, but why the second one is false?
I am assuming that $A$ is a non-trivial problem. If $A$ is trivial, then the claims cannot be correct in their current form (make sure you understand why).
Consider an NP-complete problem $B$, and consider some non-trivial problem $A \in \text{P}$. Clearly, $A\leq_p B$. If we assume that claim 2 is correct, then as $A\leq_p B$, we get that $A$ is $\text{NP}$-hard and thus for every problem $L\in \text{NP}$, it holds that $ L\leq_p A$. Hence, as $A\in \text{P}$, we get that $L\in \text{P}$ as well, and so $\text{NP}\subseteq \text{P}$. Conversely, if we assume that $\text{NP}\subseteq \text{P}$, then as every non-trivial problem is $\text{P}$-hard, we get that every non-trivial problem $A$ is $\text{NP}$-hard. Hence, claim 2 holds trivially.
So we have in total that claim 2 is correct iff $\text{P} = \text{NP}$ which is quite open.