About the fourier transform of Periodic Signal
It is not obvious what exactly OP Samuel wants to know, and so I am posting a generic answer.

Let $x(t)$ denote a finite-energy continuous-time signal by which we mean that
$\displaystyle \int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = \mathbb E_x &lt; \infty$.
It is important to understand that a finite-energy signal cannot be a periodic signal, that 
is, there is no nonzero real number $T$ for which the following equation
can hold for all time instants $t$:
$$x(t+T) = x(t)~~\text{for all}~ t, -\infty &lt; t &lt; \infty.$$
Of course, the above might hold for some, possibly even
many, values of $t$, 
but it cannot hold for all $t$.  For example, if 
$x(t) =\begin{cases}\sin(200\pi t), &amp;-1 \leq t \leq +1,\\
0, &amp;\text{otherwise,}
\end{cases}$ is a sinusoidal pulse of duration $2$ seconds
containing $200$ cycles
of a $100$ Hz sine wave, then 
$x(t+0.01) = x(t)$ for all $t$ except for $t \in (-1.01,1)$ and
$t \in(0.99,1)$.
On the other hand, we do not insist that $x(t)$ be of finite duration,
or that $x(t)$ be causal, etc. In these circumstances, $x(t)$ has
a Fourier transform $X(f)$ given by
$$X(f) 
= \int_{-\infty}^\infty x(t)\exp(-j2\pi ft)\,\mathrm dt,
~~ -\infty &lt; f &lt; \infty \tag{1}$$
where $j = \sqrt{-1}$,
and $x(t)$ can be recovered from its Fourier transform using the inverse
Fourier transform which gives
$$x(t) 
= \int_{-\infty}^\infty X(f)\exp(j2\pi ft)\,\mathrm df, 
~~-\infty &lt; t &lt; \infty.\tag{2}$$



All this, of course, is old hat to most readers, so now let us
construct a periodic signal from $x(t)$.  Let $T_0$ denote
some positive number, and define


  $$s(t) = \sum_{n=-\infty}^\infty x(t - nT_0).\tag{3}$$


Then, we claim that $s(t)$ is a signal of period $T_0$.
Note that for every choice of real number $t$, we have that
$$\begin{align}
s(t+T_0) &amp;= \sum_{n=-\infty}^\infty x(t+T_0-nT_0)\\
&amp;= \sum_{n=-\infty}^\infty x(t-(n-1)T_0) &amp;\text{and now replace}~ n-1~\text{by}~m\\
&amp;= \sum_{m=-\infty}^\infty x(t-mT_0)\\
&amp;= s(t).
\end{align}$$
But, what does one period of $s(t)$ look like? Well, for
$-T_0/2 \leq t \leq T_0/2$, $s(t)$ is not the same as $x(t)$
except in the special case when the support of $x(t)$
is a subset of $[-T_0/2,T_0/2]$. Instead, one period (what OP
Samuel calls $s_{T_0}(t)$) can be found by slicing up
the graph of $x(t)$ into pieces of duration $T_0$ centered at
integer multiples of $T_0$, moving all the pieces to be centered
at $0$, and then adding them up.  This sum, one period of the
periodic signal that we have created, is now replicated at
intervals of $T_0$ seconds all along the time axis. Thus,
we have another representation of $s(t)$, namely,


  $$s(t) = \sum_{n=-\infty}^\infty s_{T_0}(t - nT_0).
\tag{4}$$




Let us at this point introduce Dirac deltas or impulses $\delta(t)$
which have the property that for any $g(t)$ that is continuous
at $t = a$,
$$\int_{-\infty}^\infty g(t)\delta(t-a)\,\mathrm dt 
= \int_{-\infty}^\infty g(t)\delta(a-t)\,\mathrm dt = g(a).$$
Now, the convolution of $g(t)$ and $h(t)$ is a signal $(g\star h)$
whose value $(g\star h)(\tau)$ at particular time instant $\tau$ is given by
$$(g\star h)(\tau) = \int_{-\infty}^\infty g(t)h(\tau-t)\,\mathrm dt$$
and so if $h_n(t) = \delta(t-nT_0)$, we have that
$$\begin{align}
(x\star h_n)(\tau) &amp;= \int_{-\infty}^\infty x(t)h_n(\tau-t)\,\mathrm dt\\
&amp;= \int_{-\infty}^\infty x(t)\delta(\tau-t-nT_0)\,\mathrm dt\\
&amp;= \int_{-\infty}^\infty x(t)\delta((\tau-nT_0) - t)\,\mathrm dt\\
&amp;= x(\tau - nT_0).
\end{align}$$
This gives us yet another representation of the periodic signal $s(t)$,
namely,


  $$s(t)= x(t) \star \sum_{n=-\infty}^\infty \delta(t-nT_0) 
= \sum_{n=-\infty}^\infty x(t-nT_0). \tag{5}$$


It is worth reminding ourselves at this point that $s(t)$ is
a continuous-time finite amplitude signal that, despite
what it looks like in the middle sum in $(5)$ does not
contain any impulses.



Returning to Fourier transforms (bet you thought I forgot about
them, didn't you?), we come up against the uncomfortable fact
that $s(t)$ does not have a Fourier transform in the usual
sense of the word because it is not a finite-energy signal.
What it does have is a Fourier series representation,
as pointed out by Fourier himself almost two hundred years ago.
We can incorporate Fourier series into Fourier transforms
if we use impulses in the frequency domain. So, what is
the Fourier series for $s(t)$? We write
$$s(t) = \sum_{k=-\infty}^\infty c_k \exp(j2\pi kt/T_0), -\infty &lt; t &lt; \infty$$
where
$$\begin{align}
c_k &amp;= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}s(t)\exp(-j2\pi kt/T_0)\,\mathrm dt\\
&amp;= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}\sum_{n=-\infty}^\infty x(t-nT_0)
\exp(-j2\pi kt/T_0)\,\mathrm dt  &amp;\text{using}~ (3)\\
&amp;= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{-T_0/2}^{T_0/2}x(t-nT_0)
\exp(-j2\pi kt/T_0)\,\mathrm dt &amp;\text{now change variables}~ \tau = t-nT_0\\
&amp;= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{nT_0-T_0/2}^{nT_0+T_0/2}
x(\tau) \exp(-j2\pi k(\tau+nT_0)/T_0)\,\mathrm d\tau \\
&amp;= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{nT_0-T_0/2}^{nT_0+T_0/2}
x(\tau) \exp(-j2\pi k\tau/T_0)\,\mathrm d\tau
&amp;\text{now combine the integrals into one}\\
&amp;=\frac{1}{T_0}\int_{-\infty}^\infty x(\tau)\exp(-j2\pi k\tau/T_0)\,\mathrm d\tau\\
&amp;= \frac{1}{T_0}X\left(\frac{k}{T_0}\right) &amp;\text{using the definition}~ (2)
\end{align}$$

So we have the Fourier series representation
$$s(t) 
= \frac{1}{T_0}\sum_{k=-\infty}^{\infty}
X\left(\frac{k}{T_0}\right)\exp(j2\pi kt/T_0).\tag{6}$$
What should $S(f)$ be so that when it is plugged into the
inverse Fourier transform $(2)$ will give us
$$s(t) 
= \int_{-\infty}^\infty S(f)\exp(j2\pi ft)\,\mathrm df
= \frac{1}{T_0}\sum_{k=-\infty}^{\infty}
X\left(\frac{k}{T_0}\right)\exp(j2\pi kt/T_0)??$$
The answer is readily worked out to be


  $$\begin{align}
S(f) &amp;= \frac{1}{T_0}\sum_{k=-\infty}^{\infty}
X(f) \delta\left(f-\frac{k}{T_0}\right)
= \frac{1}{T_0}\sum_{k=-\infty}^{\infty}
X\left(\frac{k}{T_0}\right) \delta\left(f-\frac{k}{T_0}\right)\\
&amp;= f_0 \sum_{k=-\infty}^{\infty} X(kf_0)\delta(f-kf_0)
\end{align}$$




I will not say much about OP Samuel's "other method" since
it is comparing apples and oranges. The signal
$$x(t)\cdot \sum_{n=-\infty}^\infty \delta(t-nT_0)
= \sum_{n=-\infty}^\infty x(nT_0)\delta(t-nT_0)\tag{7}$$ is a
impulse train in the time domain (one way of
representing a sampled-data or discrete-time signal with sampling interval
$T_0$), and is very different
from the signal
$$x(t) \star \sum_{n=-\infty}^\infty \delta(t-nT_0) 
= \sum_{n=-\infty}^\infty x(t-nT_0)$$ in $(5)$ which is a continuous-time
signal containing no impulses.
There is no reason for the two to have similar or identical Fourier transforms,
and the OP's Procrustean efforts to equate both is leading him 
into a great deal
of confusion.
In fact, duality suggests that, just as the Fourier transform of
a periodic signal is a set of equally-spaced impulses (of different
amplitudes) in the frequency domain, the Fourier transform of
a set of equally-spaced impulses (of different amplitudes) in
the time domain is a periodic function in the frequency domain.
For details, see  this answer
to a different question in which the Fourier transform of $(7)$ has been
described in more depth than the hand-waving in the previous sentence.