Proving Deligne's formula If $A$ is a noetherian ring, $X = \text{Spec } A$, and $U = X - V(\mathfrak{a})$ for $\mathfrak{a}$ an ideal of $A$, then the following formula holds for any $A$-module $M$ (with associated $\mathcal{O}_X$-module $\tilde{M}$):
$$\Gamma(U,\tilde{M}) \cong \text{colim}_n \hom_A(\mathfrak{a}^n,M).$$
I've tried to prove this result by showing $\Gamma(U,\tilde{M})$ satisfies the required universal property. To that end, I defined a cocone to $\Gamma(U,\tilde{M})$ from the $\hom_A(\mathfrak{a}^n,M)$, taking
$$h_n: \hom_A(\mathfrak{a}^n,M) \to \Gamma(U,\tilde{M}), \quad h_n(f)(\mathfrak{p}) = \frac{f(x)}{x} \in M_\mathfrak{p},$$
where $\mathfrak{p} \in D(x) = V(x)^c$ for $x \in \mathfrak{a}^n$. It's easy to check that $h_n(f)$ is a well-defined section over $U$. 
The next step is to show that for any other cocone $(C, \gamma_n)$, there is a unique mediating morphism $\theta: \Gamma(U,\tilde{M}) \to C$. Finding $\theta$ should depend on realising an arbitrary $g \in \Gamma(U,\tilde{M})$ as a value of some $h_n$, but this is where I've become stuck. 
Are my maps $h_n$ wrong? If not, how to produce $g = h_n(f)$? Reductions so far: since $A$ is noetherian, $g$ can be represented as a tuple $\left ( \frac{m_1}{x_1^r}, \dots, \frac{m_k}{x_k^r} \right )$, where $m_i \in M$, the $x_i$ generate $\mathfrak{a}$, and $r$ is large.
 A: Rodney Sharp and Markus Brodmann's contribution towards helping people to understand the Deligne isomorphism is Section 20.1 of the Second Edition of their book "Local cohomology" here, but see also 2.3.2.
Brodmann, M. P.; Sharp, R. Y. Local cohomology. An algebraic introduction with geometric applications. Second edition. Cambridge Studies in Advanced Mathematics, 136. Cambridge University Press, Cambridge, 2013. xxii+491 pp.
The family $(h_n)$ of maps you suggest is alright. This family implies a map$$h: D_\mathfrak{a}(M) \to \Gamma(U, \tilde{M})$$which is the inverse of the isomorphism $\nu_{\mathfrak{a}, M}$ of 20.1.14. You will recognize this, if you consider the local family $\delta$ on page 449. (Clearly, in your case, the situation is slightly simpler, as you consider the standard affine case, whereas the hypothesis 20.1.2 concern a more general situation.)
You are actually asking whether $h$ is surjective. To prove this, you will need the fact, that for each $g \in \Gamma(U, \tilde{M})$ there is some $n \in \mathbb{N}$ such that (in the notations used in 20.1.14) $\mathfrak{a}^n g \subseteq \mathrm{Im}(\varepsilon^U_M)$. In the more general setting of the standard hypotheses 20.1.2 this is shown in 20.1.10.

Can you elaborate on how exactly to use $\mathfrak{a}^n g \subseteq \text{Im}(\varepsilon_M^U)$? Clearly I'd like to use it to construct a map $f: \mathfrak{a}^n \to M$.

Let $g \in \Gamma(U, \tilde{M}) = \tilde{M}(U)$. Then by 20.1.10 there is some $k \in \mathbb{N}$ such that $\mathfrak{a}^kg \subseteq \varepsilon_M^U(M)$. By 2.1.8(iii) we have $\text{Ker}(\varepsilon_M^U) = \Gamma_\mathfrak{a}(M)$ and so, we may identify$$\varepsilon_M^U(M) = \overline{M} := M/\Gamma_\mathfrak{a}(M).$$There is a finitely generated $R$-submodule $N \subseteq M$ such that$$(N + \Gamma_\mathfrak{a}(M))/\Gamma_\mathfrak{a}(M) = \mathfrak{a}^k g\subseteq \overline{M}.$$By Artin-Rees, there is some $t \in \mathbb{N}$ such that $\mathfrak{a}^tN \cap \Gamma_\mathfrak{a}(M) = 0$. Set $n = k + t$. Then, we get$$\mathfrak{a}^ng = (\mathfrak{a}^t N + \Gamma_\mathfrak{a}(M))/\Gamma_\mathfrak{a}(M) = (\mathfrak{a}^tN)/(\mathfrak{a}^tN \cap \Gamma_\mathfrak{a}(M)) = (\mathfrak{a}^tN)/0 = \mathfrak{a}^tN \subseteq M.$$Hence we have a map$$f \in \text{Hom}_R(\mathfrak{a}^n, M), \quad x \mapsto f(x) := xg \quad(\forall x \in \mathfrak{a}^n).$$In your notations we get:$$h_n(f)_\mathfrak{p} := {{f(x)}\over x} = {{xg}\over x} = g_\mathfrak{p} \text{ for all }\mathfrak{p} \in U \text{ and all }x \in \mathfrak{a}^n \setminus \mathfrak{p}.$$This shows that $h_n(f) = g$.
You see, that my former hint was incomplete (which I was aware of), as I did not mention the equality $\text{Ker}(\varepsilon_M^U) = \Gamma_\mathfrak{a}(M)$ of 20.1.8(iii).
Here are a few additional hints for the reading of the proof of 20.1.10 in the ordinary affine case (with $S = R$ and $T = \text{Spec}(R)$). The arguments become slightly simpler than in the book, but they base on the same ideas. Do not forget, that the situation in the book covers much more, for example the case of (multi-)projective schemes $T$...
You represent $g$ by$$\left({{m_1}\over{s_1}}, \ldots, {{m_r}\over{s_r}}\right), \text{ with }s_1, \ldots, s_r \in \mathfrak{a} \text{ and }\bigcup_{i = 1}^r (\text{Spec}(R) \setminus \text{Var}((s_i))) = U$$so that for each $\mathfrak{p} \in U := \text{Spec}(R) \setminus \text{Var}(\mathfrak{a})$ there is an $i$ such that $s_i \ni \mathfrak{p}$ and moreover, if $s_i \ni \mathfrak{p}$, then it holds in $M_\mathfrak{p}$ that $g_\mathfrak{p} = {{m_i}\over{s_i}}$ (see 20.8.1(i)).
The essential point is to show that there is some $n \in \mathbb{N}$ such that$$s_i^ng \in \varepsilon_M^U(M) \text{ for all }i = 1, 2, \ldots, r.\tag*{$(*)$}$$This is shown in the first 12 lines on page 446, on use of 20.1.19. I suggest that you read this and the proof of 20.1.9 in the book and translate it step by step to your "classical" situation in which $S = R$ and $T = \text{Spec}(R)$. I suggest this, as this reading will face you with arguments which are of basic significance if you want to learn algebraic geometry.
Let $\mathfrak{b} := \sum_{i = 1}^r Rs_i^n$. Then by $(*)$ you get$$\mathfrak{b}g \subseteq \varepsilon_M^U(M).$$It is not hard to see, that $\text{Var}(\mathfrak{b}) \subseteq \text{Var}(\mathfrak{a})$ so that $\mathfrak{a}^k \subseteq \mathfrak{b}$ for some positive integer $k$ (cf. line 14 and 15 on page 446). Therefore $\mathfrak{a}^kg \in \varepsilon_M^U(M)$.
