OK, I think I figured it out.

First let me define a few things: for any given basis $\{w_1\ldots w_k\}$ of $A\subset L^{m'}$ let $\{w'_1,\ldots w'_k\}$ be the canonical dual basis of $A'$ (the finite-dimensional dual), i-e $<w'_i,w_j>_{A',A}=\delta_{ij}$. In particular the $w'_i$'s can be seen as functions $w'_i(x)\in L^m$. Thus any linear form $v$ on $A$ can be written as $v(x)=\sum\limits_{i=1}^k\lambda_iw'_i(x)\in A'$, and for all $u(x)=\sum\limits_{i=1}^ky_iw_i(x)\in A$ we can compute the action
$$
<v,u>_{A',A}=<v,u>_{L^m,L^{m'}}=\int_{\Omega}v(x)u(x)\,dx
$$
by $L^m$ duality. For any $v\in L^m$ (and not necessarily in $A'$, this is important) we can moreover define the "projection" $P_A$ onto $A'$ by
$$
\forall u=\sum_1^ky_i w_i\in A:\qquad <P_A v,u>_{A',A}=<v,u>_{L^m,L^{m'}}=\int v(x)u(x)\,dx.
$$
Let also $\varphi_A=P_A\circ \varphi:A\to A'\subset L^m$. Now for given coefficients $\lambda_1,\ldots,\lambda_k$ the equation
$$
\forall j=1\ldots k:\qquad <\varphi(u),w_j>_{L^m,L^{m'}}=\lambda_j\hspace{2cm}(1)
$$
actually reads
$$
\text{find }u\in A\text{ such that }\varphi_A(u)=v\text{ in }A',
$$
where $v(x)=\sum_1^k\lambda_j w'_j(x)$ can be unambiguously considered either as a function in $L^m(\Omega)$ or as an element of $A'$.

**Claim 1**: given any $\lambda=\lambda_1,\ldots,\lambda_k$ equation (1) admits a unique solution $u(x)=\sum_{i=1}^k y_iw_i(x)\in A$. This comes from quite simple finite dimensional considerations by looking at the restriction $\Phi_A$ of $\Phi(u)=|u|_{L^{m'}}^{m'}$ to the subspace $A$, and noting that for all $u\in A$ the differential can be computed as
$$
\forall\, h \in A:\qquad <D\Phi_A(u),h>_{A',A}=<\varphi(u),h>_{L^m,L^{m'}}=\int \varphi(u)(x)h(x)\,dx.
$$
Please let me know if you need details (roughly, $\varphi_A$ is the gradient of a $\mathcal{C}^1(A)$ strictly convex and superlinear function $\Phi_A$, so in particular it is a bijection). This claim 1 defines in particular an inverse $\psi_A:=\varphi_A^{-1}:A'\to A$. This is just a fancy way to say that, when viewed as taking values in the dual $A'$ (by restricting the $L^m$ action to test functions in $A\subset L^{m'}$), the map $\varphi$ restricted to $A$ is invertible. This should come as no surprise since the map $\varphi(z)=z^{m'-1}$ is increasing, which is exactly what makes the PDE formally parabolic. Compared to the paper [Alt-Luckhaus] I already mentioned, this corresponds to the fact that $\varphi$ is the gradient of the convex function $\Phi$, hence a monotone vector-field (easier here since the setting is scalar, but this is really what it is).

**Claim 2:** the inverse $\psi_A$ form step 1 is locally Lipschitz. This is just a refinement of your paper, but with an extra subtlety. Let $v^1=\sum _{i=1}^k\lambda_i^1w'_i$ and $v^2=\sum_{i=1}^k \lambda_i^2w'_i$ be two given elements of $A'$, and denote the corresponding solutions to (1) by $u^1=\psi_A(v^1)$ and $u^2=\psi_A(v^2)$. Using the convexity inequality
$$
|a-b|^2\leq C(a^{m'-1}-b^{m'-1})(a-b)(|a|+|b|)^{2-m'}
$$
you get
\begin{align*}
|u^1-u^2|^2_{L^2} & \leq C\int_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)(|u^1|+|u^2|)^{2-m'}\,dx\\
& \leq C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)\,dx.
\end{align*}
In the last line I used the fact that $m'>1$ (since $m<\infty$) and monotonicity of $z\mapsto z^{m'-1}$ so that the last integrand is really pointwise nonnegative a.e $x\in \Omega$. Now because $u^1,u^2$ actually lie in the finite dimensional subspace $A$, you can rewrite this as
\begin{align*}
|u^1-u^2|^2_{L^2} & \leq C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}(\varphi(u^1)-\varphi(u^2))(u^1-u^2)\,dx\\
& =C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}(\varphi_A(u^1)-\varphi_A(u^2))(u^1-u^2)\,dx.
\end{align*}
Indeed the $\varphi(u^1),\varphi(u^2)\in L^m$ terms above "see" test functions $u^1,u^2\in A$, so their action is the same as that of their projection $\varphi_A(u^1),\varphi_A(u^2)$. Applying Young's inequality you get
$$
|u^1-u^2|^2_{L^2} \leq C(essup |u^1|+essup |u^2|)^{2-m'}|\varphi_A(u^1)-\varphi_A(u^2)|_{L^m}|u^1-u^2|_{L^{m'}},
$$
and by equivalence of the norms on $A$ you conclude that
$$
\forall \,u^1,u^2\in A:\qquad |u^1-u^2|_A\leq C(essup |u^1|+essup |u^2|)^{2-m'}\varphi_A(u^1)-\varphi_A(u^2)|_{L^m}|,
$$
which really means that the inverse $\psi_A=\varphi_A^{-1}$ is locally Lipschitz continuous if $u^1,u^2$ stay in a ball around some $u_0$. The trick was here essentially to control $|u^1-u^2|$ not by $|\varphi(u^1)-\varphi(u^)|$, which a priori takes values in the "large" $L^m$ space, but rather by their projection $|\varphi_A(u^1)-\varphi_A(u^2)|$ onto the dual $A'$, which is finite-dimensional ("small"). I guess this is what the authors had in mind in their paper, but it was well hidden and quite misleading.

**Step 3: back to the fixed point in the integral formulation.**
$$
\forall j=1\ldots k:\qquad \langle\varphi(u(t)), w_j \rangle_{L^m, L^{m'}} = \langle \varphi(u_0), w_j \rangle_{L^m, L^{m'}} + \int_0^t \langle F(s, u(s)), w_j \rangle\mathrm{d}s
$$
By the classical Lipschitz estimates in the OP the right hand-side defines, for given $u\in C(0,T;A)$ a Lipschitz continuous map $u\mapsto V[u]\in C(0,T;A')$. In other words, the above integral formulation reads:
$$
\text{find }u\in C(0,T;A)\text{ such that}\qquad \phi_a[u]=V[u]\text{ in }C(0,T;A').
$$
Now because $\psi_A=\phi_A^{-1}$ is locally Lipschitz continuous you can easily rewrite the fixed point as
$$
\text{find }u\in C(0,T;A)\text{ such that}\qquad u=\psi_A\circ V[u]\text{ in }C(0,T;A).
$$
Finally tuning the maximal time $T$ (small) and initial condition (stay in a given large ball around the "initial datum" to take care of the $essup$'s) it should be straighforward to prove that $\psi_A\circ V$ is a contraction in $C(0,T;A)$, hence the desired projected solution.