Edit: Fixed the triangles following a discussion in the comment section.
Let $X$ be a reducible algebraic variety, and let $j:U\hookrightarrow X$ be an open subvariety which is not dense in $X$. Let $i:Z \hookrightarrow X$ denote the closed complement. For convenience let us assume that all varieties are defined over $\overline{\mathbb F_p}$ for some prime number $p$, and I consider $\Lambda = \mathbb Z/\ell^h\mathbb Z$ for some $h\geq 1$ and $\ell$ a prime number different from $p$.
In an appropriate derived category $D(X,\Lambda)$, we have a distinguished triangle $$i_!i^*\Lambda \to \Lambda \to \mathrm Rj_*j^*\Lambda \xrightarrow{+1}.$$ It induces in particular a boundary map $\varphi_k: H^k(Z,\Lambda) \to H^{k+1}_c(U,\Lambda)$ for all $k\in \mathbb Z$.
Now, $j$ factors through $U \xrightarrow{j'} \overline U \xrightarrow{\iota} X$ and the closed complement of $U$ inside $\overline U$ is $i':Z' \hookrightarrow \overline U$, where $Z' := Z \times_{X} \overline U$. Similarly I have a distinguished triangle in $D(\overline U,\Lambda)$ as follows $$i'_*i^{\prime !} \Lambda \to \Lambda \to \mathrm Rj'_*j^{\prime *}\Lambda \xrightarrow{+1}.$$
It induces a boundary map $\varphi'_k :H^k(Z',\Lambda) \to H^{k+1}_c(U,\Lambda)$.
What is the relation between those two boundary maps? More precisely, is $\varphi_k$ the composition of $\varphi'_k$ with the restriction map $H^k(Z,\Lambda) \to H^k(Z',\Lambda)$ induced by the closed immersion $Z' \hookrightarrow Z$?
On LES. Here are some facts about triangulated categories, in a triangulated categories $\mathcal{T}$, if you have a distinguished triangle $$A \longrightarrow B \longrightarrow C \longrightarrow A[+1]$$ then for any $X$, you have a long exact sequence of abelian groups $$... \longrightarrow \operatorname{Hom}(A[i+1],X) \longrightarrow \operatorname{Hom}(C[i],X) \longrightarrow \operatorname{Hom}(B[i],X) \longrightarrow \operatorname{Hom}(A[i],X) \longrightarrow ...$$ where all morphism are induced by compositions (hence canonical). In the modern language, $\operatorname{Hom}(-,X)$ is a cohomological functor and you can also use $\operatorname{Hom}(X,-)$ but the direction of LES is reversed. $$...\longrightarrow \operatorname{Hom}(X,A[i]) \longrightarrow \operatorname{Hom}(X,B[i]) \longrightarrow \operatorname{Hom}(X,C[i]) \longrightarrow \operatorname{Hom}(X,A[i+1]) \longrightarrow ...$$ and afterwards, we use only the second LES.
On cohomology (with compact support). Now you should know that cohomology is a special case of ext functors, and all of them can be expressed by hom sets in appropriate derived categories. For instance, for a topological space $X$ and sheaves of abelian groups, you have $$H^n(X,\mathcal{F}) = \operatorname{Hom}_{D(\mathbf{Sh}(X))}(\mathbb{Z}_X,\mathcal{F}[n]).$$ The situation is same for etale cohomology. $$H^n(X,\mathcal{F}) = \operatorname{Hom}(\Lambda_X,\mathcal{F}[n]).$$ If moreover, you are working with schemes over a fixed base, then we can talk about cohomology with compact support. Here we assume that $X$ is a variety over an algebraically closed field $k$, denote by $\pi_X \colon X \longrightarrow \operatorname{Spec}(k)$ the structural morphism, then the cohomology with compact support is $$H^n_c(X,\mathcal{F}) = \operatorname{Hom}(\Lambda_k, (\pi_X)_!\mathcal{F}[n]).$$ Again, we can also simplify the ordinary cohomology once the base is fixed $$H^n(X,\mathcal{F}) = \operatorname{Hom}(\Lambda_k, (\pi_X)_*\mathcal{F}[n])$$ These cohomology groups are functorial in the following sense: if $f \colon X \longrightarrow Y$ be a morphism of $k$-varieties, then $\pi_Y \circ f = \pi_X$, and you have a morphism $$H^n(Y,\mathcal{G})=\operatorname{Hom}(\Lambda_k,(\pi_Y)_*\mathcal{G}[n]) \longrightarrow \operatorname{Hom}(\Lambda_k,(\pi_Y)_*f_*f^*\mathcal{G}[n]) = \operatorname{Hom}(\Lambda_k,(\pi_X)_*f^*\mathcal{G}[n]) = H^n(X,f^*\mathcal{G}).$$ And if $f$ is proper, you can do the same thing because $f_!=f_*$ $$H^n_c(Y,\mathcal{G})=\operatorname{Hom}(\Lambda_k,(\pi_Y)_!\mathcal{G}[n]) \longrightarrow \operatorname{Hom}(\Lambda_k,(\pi_Y)_!f_!f^*\mathcal{G}[n]) = \operatorname{Hom}(\Lambda_k,(\pi_X)_!f^*\mathcal{G}[n]) = H^n_c(X,f^*\mathcal{G}).$$ Localization sequences. By deriving operations $(j^*=j^!,j_*,j_!,i_*=i_!,i^!)$ from the level of sheaves, you have two localization sequences $$j_!j^*\mathcal{F} \longrightarrow \mathcal{F} \longrightarrow i_*i^*\mathcal{F} \longrightarrow j_!j^*\mathcal{F}[+1]$$ $$i_!i^!\mathcal{F} \longrightarrow \mathcal{F} \longrightarrow j_*j^*\mathcal{F} \longrightarrow i_!i^!\mathcal{F}[+1]$$ and they are duals of each other (by using duality), so in fact it is sufficient to deal with just one of them. Since you are interested in long exact sequences of cohomology with compact support, you can work with the first sequence. Apply $(\pi_X)_!$ first and then $\operatorname{Hom}(\Lambda_k,-)$ $$\operatorname{Hom}(\Lambda_k,(\pi_U)_!j^*\mathcal{F}[n]) \longrightarrow \operatorname{Hom}(\Lambda_k,(\pi_X)_!\mathcal{F}[n]) \longrightarrow \operatorname{Hom}(\Lambda_k,(\pi_Z)_!i^*\mathcal{F}[n])$$ and you know that by definition this is simply $$...\longrightarrow H^n_c(U,\mathcal{F}) \longrightarrow H^n_c(X,\mathcal{F}) \longrightarrow H^n_c(Z,\mathcal{F}) \overset{\partial}{\longrightarrow} H^{n+1}_c(U,\mathcal{F}) \longrightarrow ...$$ where the boundary map $\partial$ is given by composition $$(\Lambda_k \to (\pi_X)_!i_!i^*\mathcal{F}[n]) \longrightarrow (\Lambda_k \to (\pi_X)_!j_!j^*\mathcal{F}[n+1]).$$ Functoriality of boundary maps $\partial$. The map $\partial$ is functorial in the following sense: given cartesian squares $\require{AMScd}$ \begin{CD} U' @>>> X' @<<< Z' \\ @V{f_U}VV @V{f}VV @VVV\\ U @>>> X @<<< Z \end{CD} with $f$ proper (so $f_U,f_Z$ proper), then there is a commutative square $\require{AMScd}$ \begin{CD} H^n_c(Z,\mathcal{F}) @>{\partial}>> H^{n+1}_c(U,\mathcal{F}) \\ @VVV @VVV\\ H^n_c(Z',(f_Z)^*\mathcal{F}) @>{\partial}>> H^{n+1}_c(U',(f_U)^*\mathcal{F}) \end{CD} To see this, you can just expand the definition. In your case, you have a diagram $\require{AMScd}$ \begin{CD} U @>>> \overline{U} @<<< Z' \\ @V{=}VV @V{\iota}VV @VV{f_Z}V\\ U @>>> X @<<< Z \end{CD} which results in $\require{AMScd}$ \begin{CD} H^n_c(Z,\Lambda) @>{\partial}>> H^{n+1}_c(U,\Lambda) \\ @VVV @VV{=}V\\ H^n_c(Z',\Lambda) @>{\partial}>> H^{n+1}_c(U,\Lambda) \end{CD} by setting $\mathcal{F}=\Lambda$.