Let $M$ be a finitely-generated module over a Dedekind domain $R$. I need to show that $M = M_1 \oplus M_2$ where $M_1$ is torsion and $M_2$ is projective.

It's clear we can do this locally: indeed, for any $\mathfrak{p} \in \operatorname{Spec} R$, $M_\mathfrak{p}$ is a finitely-generated module over a DVR (a fortiori, a PID) $R_\mathfrak{p}$, so splits as a direct sum of a torsion module and a free module. Globally, there is a canonical choice for $M_1$: simply set it to be the torsion submodule of $M$. Then we find that $M/M_1$ is locally free, hence projective. Thus we have a short exact sequence
$$0 \to M_1 \to M \to M / M_1 \to 0$$
with $M_1$ torsion and $M/M_1$ projective. So far so good. It suffices now to find a map $M/M_1 \to M$ splitting the sequence above; but we know that the localised sequence
$$0 \to (M_1)_\mathfrak{p} \to M_\mathfrak{p} \to M_\mathfrak{p} / (M_1)_\mathfrak{p} \to 0$$
is split, and since every module in sight is finitely-presented, there is a $a \in R$ so that tensoring the original sequence with $R[a^{-1}]$ yields a split sequence.

Question. Is there a way for me to choose the local splittings coherently so that I can collate the maps to get a global splitting of $M$? (I'm thinking of $M$ as a coherent sheaf over $\operatorname{Spec} R$.) Or do I need to take a different approach to prove the claim?

I like to remember this as a special case of the Fubini/Tonelli theorems, where the measures are counting measure on $\mathbb{N}$ and Lebesgue measure on $\mathbb{R}$ (or $[0,\infty)$ as you've written it here).  In particular, Tonelli's theorem says if $f_n(x) \ge 0$ for all $n,x$, then $$\sum \int f_n(x) \,dx = \int \sum f_n(x) \,dx$$ without any further conditions needed.  (You can also prove this with the monotone convergence theorem.)
Then Fubini's theorem says that for general $f_n$, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$ (by Tonelli the two conditions are equivalent), then $\int \sum f_n = \sum \int f_n$.  (You can also prove this with the dominated convergence theorem.)
There may be weaker conditions that would also suffice, but these tend to work in 99% of cases.

Elaborating on request: the usual statement of Fubini's theorem goes something like this:

Let $(X,\mathcal{F}, \mu),(Y,\mathcal{G}, \nu)$ be $\sigma$-finite measure spaces, and let $g : X \times Y \to \mathbb{R}$ be measurable with respect to the product $\sigma$-algebra $\mathcal{F} \otimes \mathcal{G}$.  Suppose that $\int_X \int_Y |g(x,y)| \nu(dy) \mu(dx)$ is finite.  (Note: By Tonelli's theorem, this happens if and only if $\int_Y \int_X |g(x,y)|\mu(dx)\nu(dy)$ is finite, since both iterated integrals are equal.)   Then $$\int_X \int_Y g(x,y) \nu(dy)\mu(dx) = \int_Y \int_X g(x,y) \mu(dx) \nu(dy).$$

Let $X = \mathbb{R}$, $\mathcal{F}$ the Borel $\sigma$-algebra, and $\mu$ Lebesgue measure.  Let $Y = \mathbb{N}$, $\mathcal{G} = 2^{\mathbb{N}}$ the discrete $\sigma$-algebra, and $\nu$ counting measure.  Define $g(x,n) = f_n(x)$.  Exercise: since each $f_n$ is measurable, verify that $g$ is measurable with respect to $\mathcal{F} \otimes \mathcal{G}$.  Exercise: verify that integration with respect to counting measure is the same as summation, where the integral exists and is finite iff the sum converges absolutely.  (That is, given a sequence of real numbers $a_n$, define a function $b : \mathbb{N} \to \mathbb{R}$ by $b(n) = a_n$.  Then $\int_{\mathbb{N}} b\,d\nu = \sum_{n=1}^\infty a_n$.)
As such, the conclusion of Fubini's theorem reduces to the statement that was to be proved.
