Naive question on local cohomology

Let $X$ be a smooth, projective variety and $Z_1, Z_2$ two smooth, projective subvarieties in $X$ of the same dimension. Let $E$ be a locally free sheaf on $X$. Recall, there are natural morphims:

$$g_i:H^k_{Z_i}(E) \xrightarrow{f_i} H^k_{Z_1 \cup Z_2}(E) \to H^k(E)$$

for $i=1,2$. Let $\alpha_i \in H^k_{Z_i}(E)$ for $i=1,2$. Is it true that $g_1(\alpha_1)=g_2(\alpha_2)$ if and only if $f_1(\alpha_1)=f_2(\alpha_2)$? (If necessary assume that $Z_1 \cap Z_2$ is of codimension at least $2$ in both $Z_1$ and $Z_2$).

Any idea/reference will be most welcome.

I don't think this has much of a chance. The group $H^{k}(E)$ could easily be zero while the local cohomology is non-zero, so $g_1(\alpha_1)=g_2(\alpha_2)$ for any two classes, while if you have a single class $\alpha$ for which $f_1(\alpha)\neq 0$ then what you are hoping for doesn't work. More particularly, let $X=\mathbb P^n$, $E=\mathscr O_X$, and $Z_i\in\mathbb P$ different single points. Then for $k=n$ both $f_1$ and $f_2$ are injective, but their images only intersect in $0$, so $f_1(\alpha_1)=f_2(\alpha_2)$ only if $\alpha_1=\alpha_2=0$. My guess is that this fails more times than it is true. Notice that in this example $Z_1\cap Z_2=\emptyset$.
$$\dots \to H^{k}_{Z_1\cap Z_2}(E) \to H^k_{Z_1}(E) \oplus H^k_{Z_2}(E) \xrightarrow{\ f_1-f_2 \ } H^k_{Z_1 \cup Z_2}(E) \to H^{k+1}_{Z_1\cap Z_2}(E) \to \dots$$