The solution depends on the definition of uniform integrability you use. The most convenient one states that:
A collection $H$ of integrable functions bounded in $L^1(\Omega,\mathcal F,\mu)$ is uniformly integrable if $\int\limits_A|h|\to0$ when $\mu(A)\to0$, uniformly on $h$ in $H$, that is,
$$
\forall\varepsilon\gt0,\ \exists\eta\gt0,\ \forall A\in\mathcal F,\ \forall h\in H,\ \mu(A)\leqslant\eta\implies\int_A|h|\mathrm d\mu\leqslant\varepsilon.
$$
Then the proof is direct. First, if $\alpha H=\{\alpha h\mid h\in H\}$ and $H$ is uniformly integrable, then $\alpha H$ is bounded in $L^1$ and, if $\eta(\varepsilon,H)$ makes the implication above true for $H$ and $\varepsilon$, for each $\varepsilon$, then $\eta\left(\frac{\varepsilon}{|\alpha|},H\right)$ makes it true for $\alpha H$ and $\varepsilon$.
Second, if $H+K=\{h+k\mid h\in H,\ k\in K\}$ and $H$ and $K$ are uniformly integrable, then $H+K$ is bounded in $L^1$ and $\min\{\eta(\varepsilon,H),\eta(\varepsilon,K)\}$ makes the implication above true for $H+K$ and $2\varepsilon$.
Hence for every uniformly integrable collections $H$ and $K$ and every scalar $\alpha$ and $\beta$, $\alpha H+\beta K$ is uniformly integrable as well.
Note finally that the sequences $(f_n)_{n\in\mathbb N}$ and $(g_n)_{n\in\mathbb N}$ are uniformly integrable if and only if the collections $H=\{f_n\mid n\in\mathbb N\}$ and $K=\{g_n\mid n\in\mathbb N\}$ are uniformly integrable and that, then, $\alpha H+\beta K$ contains every function $\alpha f_n+\beta g_n$ (and many more). Hence the uniform integrability of $\alpha H+\beta K$ implies the one of $\{\alpha f_n+\beta g_n\mid n\in\mathbb N\}$.
If one wishes to use the definition that:
A collection $H$ is uniformly integrable if and only if $\int\limits_{|h|\gt t}|h|\mathrm d\mu\to0$ when $t\to+\infty$, uniformly on $h$ in $H$, that is,
$$
\forall\varepsilon\gt0,\ \exists t,\ \forall s,\ \forall h\in H,\ s\geqslant t\implies\int_{|h|\geqslant s}|h|\mathrm d\mu\leqslant\varepsilon,
$$
then the proof is similar but perhaps less convenient. Denote by $t(\varepsilon,H)$ any value of $t$ making the implication above true for $H$ and $\varepsilon$. Then $t(\varepsilon,H)$ works for $\alpha H$ and $|\alpha|\varepsilon$ hence $H$ uniformly integrable implies $\alpha H$ uniformly integrable. For the sum $H+K$, one can convince oneself of the validity of the pointwise inequality
$$
|h+k|\,\mathbf 1_{|h+k|\geqslant 2t}\leqslant 2|h|\,\mathbf 1_{|h|\geqslant t}+2|k|\,\mathbf 1_{|k|\geqslant t},
$$
and deduce that, in our notations,
$\max\{t(\varepsilon,H),t(\varepsilon,K)\}$ works for $H+K$ and $4\varepsilon$. QED.
Best Answer
We have to take $\mathbb{R}$ as the input space of $T$, as $\Phi$ can take values in all of $\mathbb{R}$. You were probably thinking about $[0;2]$ as the input space.
As @user125932 already said $T:\mathbb{R}\rightarrow\mathbb{R}$ is not uniformly continuous. As a counterexample consider a fixed $\delta>0$ and $y=x+\delta$: $$\lim_{x\rightarrow \infty}\vert x^4-(x+\delta)^4\vert=\lim_{x\rightarrow \infty}\vert 4x^3\delta+6x^2\delta^2+4x\delta^3+\delta^4\vert=\infty$$ (Another way of showing that $T$ is not uniformly continuous is basicially your way for $\Psi$ as it is the composition of $T$ and a linear function - $\Phi$ restricted to constant functions on $\mathscr{C}[0;2]$ is linear. So thats quite cool aswell! :D)
As for your question: Uniform continuity is compatible with composition and independent of the metrics used.
Consider the diagram of metric spaces and uniformly continuous maps: $$(L,d_L)\overset{f}\longrightarrow (M,d_M)\overset{g}\longrightarrow(N,d_N)$$ Since $g$ is uniformly continuous we can find for any given $\varepsilon>0$ a $\delta>0$ so that for all $x,y\in M$ we have: $$d_M(x,y)<\delta\;\Rightarrow\;d_N\left(g(x),g(y)\right)<\varepsilon$$ As $f$ is also uniformly continuous there always exists a $\delta'$ such that for all $x,y\in L$: $$d_L(x,y)<\delta'\;\Rightarrow\;d_M\left(f(x),f(y)\right)<\delta$$ So in particular for every $\varepsilon>0$ there exists a $\delta'>0$ such that for all $x,y\in L$ we have: $$d_l(x,y)<\delta'\;\Rightarrow\;d_N\left(g(f(x)),g(f(y))\right)<\varepsilon$$ and $g\circ f$ indeed is uniformly continuous. :)
Also I'd like to genuinely say that your question was extremely well written and concise, that doesn't happen that often :P