It is not difficult to find $f(n)$ for small $n$:
$$f(1 \cdot 1) = f(1)f(1) \text{ therefore } f(1) = 1;$$
$$f(3) = f(1) + f(1) + f(1) = 3;$$
$$f(5) = f(1 + 1 + 3) = 5;$$
$$f(10) = f(1 + 3 + 6) = 4 + 3f(2) \text{ and } f(10) = f(2 \cdot 5) = f(2)f(5) = 5f(2) \text{ therefore } f(2) = 2.$$
Now we use induction. Suppose that $f(n) = n$ for all $n < N$. Let us show that $f(N) = N$. Since $f$ is multiplicative we may assume that $N = p^r$ for some prime $p$. Consider several similar cases.

1) $N = 3^r$. Then
$$f(3T_{3^{r-1}}) = 3f(T_{3^{r-1}}) = 3f\left(\frac{3^{r-1}(3^{r-1}+1)}{2}\right) = 3f(3^{r-1})f\left(\frac{3^{r-1}+1}{2}\right)$$
And from the other hand
$$f(3T_{3^{r-1}}) = f\left(\frac{3^r(3^{r-1}+1)}{2}\right) = f(3^r)f\left(\frac{3^{r-1}+1}{2}\right)$$
So we conclude that $f(3^r) = 3^r$ since $f(3^{r-1}) = 3^{r-1}$ by induction hypothesis.

and
$$f(T_{s-1} + T_{s-1} + T_s) = f\left(\frac{s(3s-1)}{2}\right) = f\left(\frac{s}{2}\right)f(3s-1) = \frac{s}{2}f(p^r).$$
Hence $f(p^r) = p^r$.

3) $N = p^r$, where $p$ is an odd prime and $p^r = 3s + 1$. Similarly we have
$$\begin{aligned} f(T_{s-1} + T_s + T_s) &= \frac{s(s-1)}{2} + \frac{s(s+1)}{2} + \frac{s(s+1)}{2} = \frac{s(3s+1)}{2} = \frac{sp^r}{2} \\ &= f\left(\frac{s(3s+1)}{2}\right) = f\left(\frac{s}{2}\right)f(3s+1) = \frac{s}{2}f(p^r). \end{aligned}$$
Hence $f(p^r) = p^r$.