I don't understand what you are trying to do and here's what I would do. Consider the canonical projections $\pi_X,\pi_Y$ from $X \times Y$ to $X$ and $Y$ respectively. Let $(p,q) \in X \times Y$ and now consider the map
$$f : T_{(p,q)}(X \times Y) \to T_p X \times T_q Y$$
that sends a vector $v$ to elements $\Big(d(\pi_X)_{(p,q)}(v), d(\pi_Y)_{(p,q)}(v) \Big)$. This is linear since $d(\pi_X)_{(p,q)}$ and $d(\pi_Y)_{(p,q)}$ are both linear. On the other hand, define
$$g : T_p X \times T_q Y \to T_{(p,q)} (X \times Y)$$
that sends a pair of vectors $(v,w)$ to $d(\iota_X)_p(v) +d(\iota_Y)_q(w)$, where $\iota_X : X \to X\times Y$ sends $X$ to the slice $X \times \{q\}$ and similarly for $\iota_Y$.
Using that
\begin{align}
\pi_X \circ \iota_X =\operatorname{id}_X, \ \ \pi_Y\circ \iota_Y=\operatorname{id}_Y,
\end{align}
$\pi_Y \circ \iota_X, \ \ \pi_X\circ \iota_Y$ are constant maps and chain rule, we have
\begin{align}(f \circ g) (v, w) &= f ( d(\iota_X)_p(v) +d(\iota_Y)_q(w)) \\
&= \Big(d(\pi_X)_{(p,q)}( d(\iota_X)_p(v) +d(\iota_Y)_q(w)), d(\pi_Y)_{(p,q)}( d(\iota_X)_p(v) +d(\iota_Y)_q(w)) \Big) \\
&=\Big(d(\pi_X\circ \iota_X)_p(v) +d(\pi_X\circ\iota_Y)_q(w)), d(\pi_Y\circ\iota_X)_p(v) +d(\pi_Y\circ\iota_Y)_q(w)) \Big) \\
&= (v, w)
\end{align}
Thus $f$ is surjective. Since $T_{(p,q)}(X \times Y)$ and $ T_p X \times T_q Y$ have the same dimension, $f$ is a linear isomorphism.
Best Answer
Always start with what you know:
So now, the exercise becomes: write down the differential of the projection in the $(x_i,y_i)$ coordinate system. Can you take it from here?