From this you already know that $\displaystyle\sum\varphi_{i}\psi_{i}$ approximates $F$, so given $\epsilon\in(0,1)$, we choose polynomials $p_{i},q_{i}$ such that
\begin{align*}
\|\varphi_{i}-p_{i}\|<\dfrac{\epsilon}{nM},~~~~\|\psi_{i}-q_{i}\|<\dfrac{\epsilon}{nM},
\end{align*}
where $M=1+\|\varphi_{1}\|+\cdots+\|\varphi_{n}\|+\|\psi_{1}\|+\cdots+\|\psi_{n}\|$, then
\begin{align*}
\|\varphi_{i}\psi_{i}-p_{i}q_{i}\|&\leq\|\varphi_{i}-p_{i}\|\|\psi_{i}\|+\|p_{i}\|\|\psi_{i}-q_{i}\|\\
&<\dfrac{\epsilon\|\psi_{i}\|}{nM}+\dfrac{\epsilon(\|\varphi_{i}\|+1)}{nM}\\
&<\dfrac{2\epsilon}{n},
\end{align*}
so
\begin{align*}
\left\|\sum\phi_{i}\psi_{i}-\sum p_{i}q_{i}\right\|\leq\sum\|\phi_{i}\psi_{i}-p_{i}q_{i}\|<2\epsilon,
\end{align*}
we are done.
If one already recognizes that $C([0,1]\times[0,1])\approx C[0,1]\widehat{\otimes}C[0,1]$, the projective tensor product of copies of $C[0,1]$, the assertion will become immediate.
More generally let $V,W$ be metric spaces.
Notation: For any function $f$ and for any $T\subseteq dom(f)$ we write $f[T]=\{f(t): t\in T\}.$
Let $N=\cup_{n\in\Bbb N}\alpha_n[M].$
Let $S=(\,\alpha_{j(n)}(x_n)\,)_n$ be any sequence in $N.$ Let $(x_{i(n)})_n$ be a convergent sub-sequence of $(x_n)_n$ with limit $x\in M.$
(1). If for some $k$ we have $j(i(n))=k$ for infinitely many $n,$ then $(\,\alpha_{j(i(n))}(x_n)\,)_{[j(i(n)])=k]},$ that is, $(\,\alpha_k(x_{i(n)}\,)_{[j(i(n))=k]}$ is a subsequence of $S$ converging to $\alpha_k(x)\in N.$
(2). If $\{n: j(i(n))=k\}$ is finite for every $k,$ then we can find a subsequence $(i'(n))_n$ of the sequence $(i(n))_n$ such that $(\,j(i'(n))\,)_n$ is strictly increasing.
$(\bullet)$ The uniform convergence of $(\alpha_{j(i'(n))})_n$ to $\alpha$ then implies that $(\,\alpha_{j(i'(n))}(x_{i'(n)})\,)_n$ is a subsequence of $S$ converging to $\alpha(x).$ [Exercise for the reader.]
(3). From (1) and (2), every sequence in $N$ has a convergent subsequence, so $\overline N$ is compact.
(4). In the main Q we can replace $\Psi$ with its restriction $\Psi|_{\overline N}=^{def}F$ to the compact domain $\overline N.$ Now $F:\overline N\to W$ is uniformly continuous as both its domain and its image $F[\overline N]$ are compact. I leave it to the reader to show this implies that $(F\circ \alpha_n)_n$ converges uniformly to $F\circ \alpha.$
Hint for $(\bullet):$ I suggest a proof by contradiction, using the fact that $\alpha$ is continuous because it is a uniform limit of a sequence of
continuous functions
Best Answer
Since $F$ is uniformly continuous on the compact space $X\times T$, choose some open covers $(A_{i})_{i=1}^{n}$ and $(B_{j})_{j=1}^{m}$ for $X$ and $T$ respectively such that the oscillation of $f$ on $A_{i}\times B_{j}$ is small, say, less than $\epsilon>0$.
Then consider the partition of unity $(\varphi_{i})_{i=1}^{n}$ and $(\psi_{j})_{j=1}^{m}$ of $(A_{i})_{i=1}^{n}$ and $(B_{j})_{j=1}^{m}$ respectively. For each $1\leq i\leq n$ and $1\leq j\leq m$, pick an $a_{i,j}=f(t_{i},s_{j})$, $t_{i}\in A_{i}$, $s_{j}\in B_{j}$, then argue that $\displaystyle\sum_{1\leq i\leq n,1\leq j\leq m}a_{i,j}\varphi_{i}\psi_{j}$ approximates $f$ in the uniform norm.
The $f_{k}$ and $g_{k}$ are chosen to be suitable rearrangement of $a_{i,j}\varphi_{i}\psi_{j}$.
Edit:
The partition of unity is such that $\displaystyle\sum_{1\leq i\leq n}\varphi_{i}=1$ and $\displaystyle\sum_{1\leq j\leq m}\psi_{j}=1$ and hence \begin{align*} f(s,t)=\sum_{1\leq i\leq n, 1\leq j\leq m}f(s,t)\varphi_{i}(s)\psi_{j}(t), \end{align*} then \begin{align*} &\left|f(s,t)-\sum_{1\leq i\leq n,1\leq j\leq m}a_{i,j}\varphi_{i}\psi_{j}\right|\\ &\sum_{1\leq i\leq n, 1\leq j\leq m}|f(s,t)-a_{i,j}|\varphi_{i}(s)\varphi_{j}(t)\\ &<\epsilon\cdot\sum_{1\leq i\leq n, 1\leq j\leq m}\varphi_{i}(s)\varphi_{j}(t)\\ &=\epsilon. \end{align*}