Let $\Lambda := \mathbb{Z}/m\mathbb{Z}$, $X$ a scheme, $i : Y\hookrightarrow X$ a closed immersion, $j : U := X-Y\hookrightarrow X$ the open immersion of the complement, and $\mathcal{F}$ an etale sheaf of $\Lambda$-modules on $X$.
In Freitag-Kiehl p106 (Section 10 of Chapter 1), he describes the functors
$$\Gamma_Y : (\Lambda\text{-mod})(X)\rightarrow\text{Ab}\qquad \Gamma_Y(\mathcal{F}) := Ker(\mathcal{F}(X)\rightarrow\mathcal{F}(U))$$
and $$\textbf{L}_Y : (\Lambda\text{-mod})(X)\rightarrow(\Lambda\text{-mod})(X)\qquad \textbf{L}_Y(\mathcal{F}) := Ker(\mathcal{F}\rightarrow j_*j^*\mathcal{F})$$ They are both left exact, and he writes $H_Y^i(X,\mathcal{F}) = R^i\Gamma_Y(\mathcal{F})$.
On p107, as shown here, he says that "we have the usual long exact sequences":
$$0\rightarrow\textbf{L}_Y(\mathcal{F})\rightarrow\mathcal{F}\rightarrow j_*j^*\mathcal{F}\rightarrow R^1\textbf{L}_Y(\mathcal{F})\rightarrow 0\qquad\qquad(*)$$ and $R^ij_*j^*\mathcal{F}\cong R^{i+1}\textbf{L}_Y(\mathcal{F})$ for $i\ge 1$.
I'm pretty confused how he gets this. I can't see a short exact sequence for which $(*)$ is the first few terms of the associated long exact sequence.
Okay, so as usual the stacks project comes to the rescue.
The usual way of producing long exact sequences (say, to the right) is by applying a left exact functor $F$ to a short exact sequence in some abelian category, from which we get the long exact sequence with the higher derived functors $R^iF$. However, another way of saying this is that a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ in an abelian category $\mathcal{A}$ yields a distinguished triangle in the derived category $D(\mathcal{A})$, namely, this is the triangle represented by $A[0]\rightarrow B[0]\rightarrow C[0]\rightarrow A[1]$ where the last map is obtained using mapping cones, and where $A[0]$ denotes the complex supported in degree 0 with value $A$, and similarly for $B[0],C[0]$.
This, combined with the result that "taking 0-th cohomology" is a homological functor tells us that we have a long exact sequence in $\mathcal{A}$: $$\cdots \rightarrow H^0(A[0])\rightarrow H^0(B[0])\rightarrow H^0(C[0])\rightarrow H^0(A[1])\rightarrow H^0(B[1])\rightarrow H^0(C[1])\rightarrow\cdots$$ In the usual setting, all this amounts to saying is that $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence, which says nothing new. However, if $F:\mathcal{A}\rightarrow\mathcal{B}$ is a left exact functor between abelian categories where $\mathcal{A}$ has enough injectives, then the nontrivial statement is that since derived functors preserve distinguished triangles, $RF(A[0])\rightarrow RF(B[0])\rightarrow RF(C[0])\rightarrow RF(A[1])$ is also a distinguished triangle in $D(\mathcal{B})$, and hence we obtain a long exact sequence $$\cdots\rightarrow H^0(RF(A[0]))\rightarrow H^0(RF(B[0]))\rightarrow H^0(RF(C[0]))\rightarrow H^0(RF(A[1]))\rightarrow\cdots$$ Unraveling definitions, replacing $A[0],B[0],C[0]$ with injective resolutions $I^\bullet, J^\bullet,K^\bullet$ supported in nonnegative degrees, this is precisely the long exact sequence $$0\rightarrow H^0(I^\bullet)\rightarrow H^0(J^\bullet)\rightarrow H^0(K^\bullet)\rightarrow H^1(I^\bullet)\rightarrow H^1(J^\bullet)\rightarrow\cdots$$ of derived functors.
The description above shows that ultimately every long exact sequence comes from a distinguished triangle in the derived category. The usual long exact sequence is produced by applying a derived functor to a short exact sequence. However, in the problem at hand, a distinguished triangle will be produced in another way.
Namely, I claim that there is a distinguished triangle in the derived category given by: $$R\textbf{L}_Y(\mathcal{F}[0])\rightarrow \mathcal{F}[0]\rightarrow R(j_*j^*)(\mathcal{F}[0])\rightarrow R\textbf{L}_Y(\mathcal{F}[0])[1]\qquad\qquad(**)$$ Indeed, we may define this triangle to be constructed as follows. First, let $\mathcal{F}[0]\cong I^\bullet$ be an injective resolution, then the sequence $$0\rightarrow\textbf{L}_Y(I^\bullet)\rightarrow I^\bullet\rightarrow j_*j^*I^\bullet\rightarrow 0$$ is exact (in the abelian category of complexes) since the second map is surjective (injective modules being flasque). Thus, this gives a triangle in the derived category. Since $I^\bullet$ is a complex of injectives, the triangle can be written as $$R\textbf{L}_Y(I^\bullet)\rightarrow I^\bullet\rightarrow R(j_*j^*)(I^\bullet)\rightarrow R\textbf{L}_Y(I^\bullet)[1]\qquad (***)$$ and since derived functors respect quasi-isomorphisms, we define the triangle $(**)$ to be obtained from $(***)$ via the canonical isomorphisms $$R\textbf{L}_Y(I^\bullet)\cong R\textbf{L}_Y(\mathcal{F}[0]), \quad I^\bullet\cong \mathcal{F}[0], \quad Rj_*j^*I^\bullet\cong Rj_*j^*\mathcal{F}[0]$$
Thus, $(**)$ is a distinguished triangle, which gives precisely the long exact sequence "$(*)$" alluded to on p107, and the "dimension shifting isomorphism" follows immediately from the fact that $\mathcal{F}[0]$ is supported in degree 0.