I have solved problem $7W$ in a particular case when $\varphi$ is uniformly continuous and I wonder if there are any counterexamples when uniform continuity isn't satisfied.
7.W. If $f_n$ converges to $f$ in $L^p$ on an arbitrary measure space and $\varphi$ is continuous and satisfies $|\varphi(t)|\leq K|t|$ for all $t\in \mathbb{R}$, then $\varphi \circ f_n$ converges to $\varphi \circ f$ in $L^p$. Conversely if $\varphi$ does not satify $|\varphi(t)|\leq K|t|$, there exists a measure space and a sequence $f_n$ which converges in $L^p$ to $f$, but such that $\varphi\circ f_n$ does not converge in $L^p$ to $\varphi\circ f$.
Before you read my answer below, I wanted to share a little bit of theory which is needed, but you may skip these because they are fairly well known:
1. Convergence in $L^p$ implies convergence in measure.
2. Convergence in measure implies there is a subsequence that converges pointwise almost everywhere and almost uniformly.
3. Almost uniform convergence implies convergence in measure.
We finally prove Bartles 7W but we ask $\varphi$ to be uniformly continous.
$(\Rightarrow)$ We have $f_n\xrightarrow[]{L^p} f$ so $f_n\xrightarrow[]{\mathcal{M}} f$. But this implies there is a subsequence such that $f_{n_k}\xrightarrow[]{} f$ pointwise almost everywhere and also almost uniformly. Because $\varphi$ is continuous, we have $\varphi\circ f_{n_k}\xrightarrow[]{} \varphi \circ f$ pointwise almost everywhere and because $\varphi$ is uniformly continuous we have $\varphi\circ f_{n_k}\xrightarrow[]{} \varphi \circ f$ also almost uniformly.
We prove $\varphi\circ f_{n_k}\xrightarrow[]{L^p} \varphi \circ f$ using Vitali's Theorem:
1.1.$\varphi\circ f_{n_k}\xrightarrow[]{\mathcal{M}} \varphi \circ f$
This follows from the fact almost uniform convergence implies convergence in measure.
1.2.$\exists \delta(\varepsilon)>0$ such that if $\mu(E)<\delta$, $\int |\varphi\circ f_{n_k}|d\mu<\varepsilon^p$ for all $k\in \mathbb{N}$.
This follows from the fact $f_n$ satisfies a similar property property with $\varepsilon'=\epsilon/K$:
$$\int_E|\varphi\circ f_{n_k}|^pd\mu\leq K^p \int_E|f_{n_k}|^p d\mu <K^p\epsilon'^p=\epsilon^p$$
1.3.$\exists E_\varepsilon$ such that $\mu(E_\varepsilon)<\infty$ and $\int_{E_\varepsilon^C} |\varphi \circ f_{n_k} |^p d\mu<\epsilon^p$ for all $k\in \mathbb{N}$
This follows from the fact $f_n$ satisfies this property with $\varepsilon'=\epsilon/K$:
$$\int_{E_\varepsilon^C}|\varphi\circ f_{n_k}|^pd\mu\leq K^p \int_{E_\varepsilon^C}|f_{n_k}|^p d\mu <K^p\epsilon'^p=\epsilon^p$$
This finishes the proof that $\varphi \circ f_{n_k}$ converges to $\varphi \circ f$ in $L_p$.If $\varphi \circ f_n $ didn’t converge to $\varphi \circ f$ in $L^p$, we could extract a subsequence which wouldn’t take any subsequence convergent to $\varphi \circ f$ in contradiction to what we proved.
$(\Leftarrow)$ For every natural number there $\exists t_n$ such that $\varphi(t_n)> n|t_n|$.
Case I:$t_n\not=0 \: ,\:\forall n$. We take $\mathbb{N}$ with the partition sigma algebra and also discrete measure $\mu(E)=\sum_{n\in E} \frac{1}{|t_n|^p n^2}$. Let $f(n)=|t_n|$. It is easilly seen that:
$$\int|\varphi \circ f|^pd\mu=\sum_n|\varphi\circ f(n)|^p\mu(n)\geq\sum_n n^p|t_n|^p \mu(n)\geq \sum_n\frac{1}{n}=\infty $$$$\int |f|^pd\mu=\sum_n|t_n|^p\mu(n)=\sum_n\frac{1}{n^2}=\frac{\pi^2}{6}<\infty$$
Case II:$\exists t_n$ such that $t_n=0$. We take $\mathbb{N}$ with the partition sigma algebra and also the counting measure. $|\varphi(0)|>0$. We take $f\equiv 0$.
$$\int|\varphi\circ f|^p d\mu=\sum_n |\varphi(0)|^p=\infty$$$$\int|f|^p d\mu =0 <\infty$$
In both cases, we define $f_n=(1-1/n)f$. Clearly they are dominated by $|f|$ which is in $L_p$, and so, by the dominated convergence theorem $f_n \xrightarrow[]{L_p} f$. However, $\varphi \circ f_n$ cannot converge to $\varphi \circ f$ because this last function is not in $L^p$.