To give a motivation as to my answer, let me first write out the proof that $Y\subseteq conv(x_1/t_1,\ldots,x_K/t_K)$. In this case, there exist $x\in\mathbb R^n,t>0$ and non-negative numbers $\lambda_1,\ldots,\lambda_K$ such that $y=\frac xt$ and
$$
\sum\limits_{i=1}^K\lambda_ix_i=x,\qquad\sum\limits_{i=1}^K\lambda_it_i=t,\qquad\sum\limits_{i=1}^K\lambda_i=1,
$$
by definition. Observe that
$$
y=\frac xt=\sum\limits_{i=1}^K\frac{\lambda_ix_i}t=\sum\limits_{i=1}^K\left(\frac{t_i}{t_i\sum\limits_{j=1}^K\lambda_jt_j}\right)\lambda_ix_i=\sum\limits_{i=1}^K\left(\frac{\lambda_it_i}{\sum\limits_{j=1}^K\lambda_jt_j}\right)\frac{x_i}{t_i},
$$
and since
$$
\sum\limits_{i=1}^K\frac{\lambda_it_i}{\sum\limits_{j=1}^K\lambda_jt_j}=1,
$$
it follows $y\in conv(x_1/t_1,\ldots,x_K/t_K)$.
Conversely, if $z\in conv(x_1/t_1,\ldots,x_K/t_K)$, then there exist non-negative numbers $\lambda_1,\ldots,\lambda_K$ such that $z=\sum\limits_{i=1}^K\lambda_i\frac{x_i}{t_i}$, and $\sum\limits_{i=1}^K\lambda_i=1$. The problem is resolved as soon as we have a list of non-negative numbers $\delta_1,\ldots,\delta_K$ such that $\sum\limits_{i=1}^K\delta_i=1$ and
$$
\lambda_i=\frac{\delta_i t_i}{\sum\limits_{j=1}^K\delta_jt_j},\quad\text{for each }i=1,2,\ldots,K.
$$
This is because, in this case, we can write
$$
z=\sum\limits_{i=1}^K\left(\frac{\delta_i t_i}{\sum\limits_{j=1}^K\delta_jt_j}\right)\frac{x_i}{t_i}=\frac{\sum\limits_{i=1}^K\delta_ix_i}{\sum\limits_{j=1}^K\delta_jt_j}=:\frac{x}t,
$$
and, by construction, $(x,t)\in conv(S)$. So the problem is to find the list of numbers $\delta_i$'s. However, this is a linear algebra problem ($K$ variables with $K+1$ equations, one of which is non-homogeneous), and one can construct the $\delta_i$'s directly from the $\lambda_i$'s. For instance, for any $c>0$, if we let
$$
\delta_i':=c\frac{\lambda_i}{t_i},\quad\text{for each }i=1,2,\ldots,K,
$$
then by multiplying through $t_i$ in each of the above definitions and then adding up the equations, we get:
$$
c=\sum\limits_{j=1}^K\delta_j't_j,
$$
and upon letting $\delta_i=\frac{\delta_i'}{\sum\limits_{j=1}^K\delta_j'}$, it follows the list $\delta_1,\ldots,\delta_K$ is of the desired form.