The derivation below of the relation between samples of a function, $s(t)$, and samples of the Fourier transform of the same function, $\hat s(f)$, can be found in this wiki article. $$\begin{align} \sum_{k=-\infty}^{\infty} \hat s(\nu + k/T) &= \sum_{k=-\infty}^{\infty} \mathcal{F}\left \{ s(t)\cdot e^{-i 2\pi\frac{k}{T}t}\right \}\\ &= \mathcal{F} \bigg \{s(t)\underbrace{\sum_{k=-\infty}^{\infty} e^{-i 2\pi\frac{k}{T}t}}_{T \sum_{n=-\infty}^{\infty} \delta(t-nT)}\bigg \} = \mathcal{F}\left \{\sum_{n=-\infty}^{\infty} T\cdot s(nT) \cdot \delta(t-nT)\right \}\\ &= \sum_{n=-\infty}^{\infty} T\cdot s(nT) \cdot \mathcal{F}\left \{\delta(t-nT)\right \} = \sum_{n=-\infty}^{\infty} T\cdot s(nT) \cdot e^{-i 2\pi nT \nu}. \end{align}$$
What is the justification behind the replacement of $s(t)$ by $s(nT)$ in the second line of this derivation?
we know that multiplication by shifted delta function means sampling value of function at that instant only i.e,
$f(t)\times \delta(t-t_{0})=f(t_{0})\times\delta(t-t_{0})$ (sifting property)
now, coming to your second line
$= \mathcal{F} \bigg \{s(t)\underbrace{\sum_{k=-\infty}^{\infty} e^{-i 2\pi\frac{k}{T}t}}_{T \sum_{n=-\infty}^{\infty} \delta(t-nT)}\bigg \}$
$= \mathcal{F}\left \{\sum_{n=-\infty}^{\infty} T\cdot s(t) \cdot \delta(t-nT)\right \}\\$
$ = \mathcal{F}\left \{\sum_{n=-\infty}^{\infty} T\cdot s(nT) \cdot \delta(t-nT)\right \}\\$ (replacement of $s(t)$ by $s(nT)$ is done by using sifting property with $t_{0}=nT$ )