Although there is already an accepted answer this example may still be of some interest:

Example Let $S=k[x,y,z]$ with the usual grading and consider the homomorphism of graded modules $$ S(-2)\to S(-1)\oplus S(-1) $$ given by taking the generator of $S(-2)$ to $(y,x)$. The cokernel of this morphism is the usual way to generate the ideal $(x,y)\subset S$. Considering the associated morphisms of locally free sheaves on $\mathbb P^2$ we get a short exact sequence as in the question: $$ 0\to \mathscr O_{\mathbb P^2}(-2) \to \mathscr O_{\mathbb P^2}(-1)\oplus \mathscr O_{\mathbb P^2}(-1) \to \mathfrak m\to 0, $$ where $\mathfrak m$ is the ideal of the point $[0:0:1]\in\mathbb P^2$. Next twist this sequence by $\mathscr O_{\mathbb P^2}(2)$ to get $$ 0\to \mathscr O_{\mathbb P^2} \to \mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1) \to \mathfrak m\otimes \mathscr O_{\mathbb P^2}(2)\to 0. $$ Now take the section $1\in H^0(\mathscr O_{\mathbb P^2})$ and observe that its image in $\mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1)$ is $(y,x)$ and hence its zero set is the point $[0:0:1]\in\mathbb P^2$. $\square$

Observe that this is not a very special example and similar phenomena appear on any (smooth projective) variety of dimension at least $2$. For simplicity assume $X$ is defined over an algebraically closed field $k$. Let $x\in X$ be a smooth closed point and consider a locally free resolution of the sheaf $k$ supported at $x$. Since $X$ is smooth at $x$, we may choose a finite resolution and choose one of minimal length. Let $\phi:\mathscr F\to \mathscr E$ be the last morphism in the resolution and observe that $\phi$ is injective and the cokernel of $\phi$, say $\mathscr Q$, is a torsion-free sheaf that is not locally free. So we have a short exact sequence: $$ 0\to \mathscr F \to \mathscr E \to \mathscr Q \to 0 $$ Suppose $s\in H^0(X,\mathscr F)$ and that $y\in X$ is in the zero locus of $s$. This is equivalent to saying that $s\in H^0(X,\mathscr F\otimes \mathfrak n)$ where $\mathfrak n$ is the ideal sheaf of $y\in X$. (Note that since $\mathscr F$ is locally free, $\mathscr F\otimes \mathfrak n\subseteq \mathscr F$. By functoriality it follows that when considered as a section of $\mathscr E$ (using the notation in the question), $s_E\in H^0(X,\mathscr E\otimes \mathfrak n)$ and hence $y$ is still in the zero section of $s$. In other words:

Good news: The zero locus of $s_E$ contains the zero locus of $s$.

However, the above example shows that

Bad news: In general the zero locus of $s_E$ is strictly larger than the zero locus of $s$.

I think it is reasonable to expect that this will be the typical behaviour. I leave it to the reader to figure out some conditions that ensure this.

Anyway, this is at the end is not surprising, since this happens all the time if we do not require the quotient to be torsion sheaf: If for instance a locally free sheaf of rank $1$ is contained in another locally free sheaf of rank $1$, that means that the corresponding divisors contain each other and if it is a proper containment, then one of the divisors is strictly bigger and hence if there is any non-zero section of the smaller sheaf, then its zero locus will be strictly smaller than that of the larger sheaf. The only "trick" here is to make the quotient torsion free, but with allowing higher rank and higher dimension it can happen.

Although there is already an accepted answer this example may still be of some interest:

Example Let $S=k[x,y,z]$ with the usual grading and consider the homomorphism of graded modules $$ S(-2)\to S(-1)\oplus S(-1) $$ given by taking the generator of $S(-2)$ to $(y,x)$. The cokernel of this morphism is the usual way to generate the ideal $(x,y)\subset S$. Considering the associated morphisms of locally free sheaves on $\mathbb P^2$ we get a short exact sequence as in the question: $$ 0\to \mathscr O_{\mathbb P^2}(-2) \to \mathscr O_{\mathbb P^2}(-1)\oplus \mathscr O_{\mathbb P^2}(-1) \to \mathfrak m\to 0, $$ where $\mathfrak m$ is the ideal of the point $[0:0:1]\in\mathbb P^2$. Next twist this sequence by $\mathscr O_{\mathbb P^2}(2)$ to get $$ 0\to \mathscr O_{\mathbb P^2} \to \mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1) \to \mathfrak m\otimes \mathscr O_{\mathbb P^2}(2)\to 0. $$ Now take the section $1\in H^0(\mathscr O_{\mathbb P^2})$ and observe that its image in $\mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1)$ is $(y,x)$ and hence its zero set is the point $[0:0:1]\in\mathbb P^2$. $\square$

Observe that this is not a very special example and similar phenomena appear on any (smooth projective) variety of dimension at least $2$. For simplicity assume $X$ is defined over an algebraically closed field $k$. Let $x\in X$ be a smooth closed point and consider a locally free resolution of the sheaf $k$ supported at $x$. Since $X$ is smooth at $x$, we may choose a finite resolution and choose one of minimal length. Let $\phi:\mathscr F\to \mathscr E$ be the last morphism in the resolution and observe that $\phi$ is injective and the cokernel of $\phi$, say $\mathscr Q$, is a torsion-free sheaf that is not locally free. So we have a short exact sequence: $$ 0\to \mathscr F \to \mathscr E \to \mathscr Q \to 0 $$ Suppose $s\in H^0(X,\mathscr F)$ and that $y\in X$ is in the zero locus of $s$. This is equivalent to saying that $s\in H^0(X,\mathscr F\otimes \mathfrak n)$ where $\mathfrak n$ is the ideal sheaf of $y\in X$. (Note that since $\mathscr F$ is locally free, $\mathscr F\otimes \mathfrak n\subseteq \mathscr F$. By functoriality it follows that when considered as a section of $\mathscr E$ (using the notation in the question), $s_E\in H^0(X,\mathscr E\otimes \mathfrak n)$ and hence $y$ is still in the zero section of $s$. In other words:

Good news: The zero locus of $s_E$ contains the zero locus of $s$.

However, the above example shows that

Bad news: In general the zero locus of $s_E$ is strictly larger than the zero locus of $s$.

I think it is reasonable to expect that this will be the typical behaviour. I leave it to the reader to figure out some conditions that ensure this.

Anyway, at the end this is not surprising, since this happens all the time if we do not require the quotient to be torsion sheaf: If for instance a locally free sheaf of rank $1$ is contained in another locally free sheaf of rank $1$, that means that the corresponding divisors contain each other and if it is a proper containment, then one of the divisors is strictly bigger and hence if there is any non-zero section of the smaller sheaf, then its zero locus will be strictly smaller than that of the larger sheaf. The only "trick" here is to make the quotient torsion free, but with allowing higher rank and higher dimension it can happen.