How to calculate 15(7)? 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 $k(7)$.

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.
