Let $X$ be a separated scheme of finite type over an algebraically closed field $k$ of characteristic $p>0$, possibly not equidimensional. Assume that $X$ has only two irreducible components $X_1$ and $X_2$, and that each $X_i$ is a nonsingular complete variety. Assume further that $X_1$ and $X_2$ meet transversally at $Y := X_1 \cap X_2$. In particular, $Y$ is nonsingular, and the singular locus of $X$ consists precisely of $Y$. Let $U_i := X \setminus X_i$ denote the open complement of $X_i$. Finally, fix $\ell$ a prime different from $p$. We will write $H^i(\,\cdot\,) := H^i(\,\cdot\,,\mathbb Q_{\ell})$ for the $\ell$-adic cohomology groups of the various varieties above.
Recall the existence of the long exact sequence of cohomology with compact support given a closed subvariety of $X$ and its open complement. Let me form the following diagram.
$$\begin{array}{ccc} \vdots & & \vdots \\ \downarrow & & \uparrow \\ H_c^i(U_1) & \rightarrow & H^i(X_2) \\ \downarrow & & \uparrow \\ H^i(X) & = & H^i(X) \\ \downarrow & & \uparrow \\ H^i(X_1) & \leftarrow & H^i_c(U_2)\\ \downarrow & & \uparrow \\ \vdots & & \vdots \end{array}$$
Here, the columns are the LES for the triples $(U_1,X,X_1)$ and $(U_2,X,X_2)$. The horizontal maps come from the LES for the triples $(U_1,X_2,Y)$ and $(U_2,X_1,Y)$, after noticing that $U_i = X \setminus X_i = X_{i+1} \setminus Y$.
Is this diagram commutative? Of course, I am well aware that the LES is functorial with respect to the pair $(U,Z)$ (the open subset $U$ and its closed complement $Z$), but unless I'm mistaken my question does not seem to be just an application of this property.
For any $k$-variety $\pi_Z \colon Z\longrightarrow \operatorname{Spec}(k)$, let's write $D(Z)$ for $D^b_c(Z,\overline{\mathbb{Q}}_l)$ whose unit object is denoted by $\mathbb{1}_Z$, then $H^i(Z) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_Z)_*\mathbb{1}_Z[i])$ and $H^i_c(Z) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_Z)_!\mathbb{1}_Z[i])$.
Your long exact sequences should involve cohomology with compact support. By this I mean it should be $H^i_c(U_1) \longrightarrow H^i_c(X) \longrightarrow H^i_c(X_1)$ and by properness $H^i_c(X_1) = H^i(X_1)$.
Let $v_1 \colon U_1 = X_2 \setminus Y \hookrightarrow X_2,t_2 \colon X_2 \hookrightarrow X$ be inclusions. Then the (co)unit morphisms $(t_2 \circ v_1)_!\mathbb{1}_{U_1}\to \mathbb{1}_X$, $\mathbb{1}_X \to (t_2)_!\mathbb{1}_{X_2}$ and $(v_1)_!\mathbb{1}_{U_1} \to \mathbb{1}_{X_2}$, respectively induce morphisms (simply by taking composition)
$$H^i_c(U_1) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{U_1})_!\mathbb{1}_{U_1}[i]) \to H^i_c(X) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{X})_!\mathbb{1}_{X}[i]) $$ $$H^i_c(X) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{X})_!\mathbb{1}_{X}[i]) \to H^i_c(X_2) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{X_2})_!\mathbb{1}_{X_2}[i])$$ $$H^i_c(U_1) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{U_1})_!\mathbb{1}_{U_1}[i]) \to H^i_c(X_2) = \operatorname{Hom}_{D(k)}(\mathbb{1}_k,(\pi_{X_2})_!\mathbb{1}_{X_2}[i])$$ Now it is necessary to have a factorization $H^i_c(U) \to H^i_c(X) \to H^i_c(X_2)$ thanks to functoriality $$(\pi_{U_1})_!\mathbb{1}_{U_1} = (\pi_X)_!(t_2 \circ v_1)_!\mathbb{1}_{U_1} \longrightarrow (\pi_X)_!\mathbb{1}_X \to (\pi_X)_!(t_2)_!\mathbb{1}_{X_2}= (\pi_{X_2})_!\mathbb{1}_{X_2}.$$