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I want to consider the value of the 'function' $15$ at the point $(7)$ of $\text{Spec}(\Bbb Z)$.

So we consider $15\in\Bbb Z$ over the composition: $$\Bbb Z\to \Bbb Z/(7)\to k(7)$$

Where the last term is the residue field at $7$.

So we have $15\mapsto [1]\in\Bbb Z/(7)$, and I take it that $k(7)=\Bbb Z_{(7)}/(7)\Bbb Z_{(7)}$ (where $\Bbb Z_{(7)}$ is the localization at prime ideal $(7)$). Where this just consists of all fractions $i/j$ such that $i,j\in \Bbb Z$ and $7$ does not divide $i$ or $j$.

I have no idea where to send $[1]\in \Bbb Z/(7)$ to in $k(7)$.

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    $k(7)=\operatorname {Frac}(\mathbb Z/(7))=\mathbb Z/(7)$, so that $ \Bbb Z/(7)\to k(7)=\mathbb Z/(7)$ is the identity map and thus $[1]\mapsto [1]$ – 2017-02-12
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    Wait I messed up my comment and was writing elements of the maximal ideal. Let me think about this – 2017-02-12
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    @GeorgesElencwajg $1,2,3,4,5,6,8,9,10,11,12,13,15$ are all distinct in $k(7)$ but are not distinct in $\Bbb Z/(7)=\Bbb Z/7\Bbb Z$ right?(Unless $\Bbb Z/(7)$ means something else entirely and I've confused this notation) – 2017-02-12
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    Perhaps I see. $1/2=4/1$ for example, since $0=s''(rs'-sr')=s''(1-8)=s''(-7)\in (7)\Bbb Z_{(7)}$ so $[4]=[1/2]$. In particular $\Bbb Z/(7)$ is a finite field $\Bbb F_7$ so it's field of fractions is itself, and $[8]=[1]$ as usual. – 2017-02-12

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Let's back up a bit. In general, if $A$ is a commutative ring and $P\subset A$ is a prime ideal, the residue field $k(P)=A_P/PA_P$ has a map $f:A/P\to k(P)$ defined by $f(a+P)=\frac{a}{1}+PA_P$. In other words, we take the canonical inclusion map $A\to A_P$ (taking $a\in A$ to the fraction $\frac{a}{1}$), and compose it with the quotient map $A_P/PA_P$. The resulting homomorphism $A\to A_P/PA_P$ vanishes on $P$, so it induces a homomorphism $A/P\to A_P/PA_P$.

So, in your case, you just follow this reciple in the case $A=\mathbb{Z}$ and $P=(7)$. The coset of $1$ in $\mathbb{Z}/(7)$ will map to the coset of $\frac{1}{1}$ in $\mathbb{Z}_{(7)}/(7)\mathbb{Z}_{(7)}$.

It should be remarked, though, that in this case the map in question is an isomorphism. Indeed, more generally, the map $A/P\to A_P/PA_P$ just realizes $A_P/PA_P$ as the field of fractions of the domain $A/P$. When $P$ is a maximal ideal (as it is in this case), $A/P$ is already a field.