The following proposition provides only a partial response to the latest edit.
Proposition. There does not exist an $ f: \Bbb{R} \to \Bbb{R} $ that fits the stronger requirement reflected in the OP’s latest edit if we assume $ g $ to be continuously differentiable.
Proof
As $ g $ is, by assumption, surjective and differentiable everywhere, there exists an $ a \in \Bbb{R} $ such that $ g'(a) \neq 0 $. Then by the further assumption that $ g': \Bbb{R} \to \Bbb{R} $ is continuous, the Inverse Function Theorem says that we can find an open interval $ I $ containing $ a $ satisfying the following:
- $ g[I] $ is an open interval containing $ g(a) $.
- $ g|_{I}: I \to g[I] $ is invertible.
- $ (g|_{I})^{-1}: g[I] \to I $ is differentiable.
We can now write $ f|_{g[I]} $ as the composition of two differentiable functions:
$$
f|_{g[I]} = (f \circ g) \circ (g|_{I})^{-1}.
$$
By the Chain Rule, $ f|_{g[I]} $ is differentiable on $ g[I] $, so we conclude that $ f $ is differentiable on some open interval at least (if $ g' $ is continuous). $ \quad \blacksquare $
Latest Edit
This latest edit, although it does not answer the question in its entirety, shows that $ f $ cannot be badly behaved everywhere.
Theorem. Let $ f: \Bbb{R} \to \Bbb{R} $ and $ g: \Bbb{R} \to \Bbb{R} $ be surjective functions such that both $ g $ and $ f \circ g $ are differentiable everywhere. Then there are uncountably many $ a \in \Bbb{R} $ such that $ f $ possesses a one-sided derivative at $ g(a) $.
Proof
As $ g $ is not constant on $ \Bbb{R} $, there exist by Darboux’s Theorem uncountably many $ a \in \Bbb{R} $ such that $ g'(a) \neq 0 $. Fix such an $ a $, and choose a $ \Delta > 0 $ such that
$$
\forall h \in [- \Delta,\Delta] \setminus \{ 0 \}: \quad
\frac{g(a + h) - g(a)}{h} \neq 0.
$$
In particular, $ g(a + h) - g(a) \neq 0 $ for all $ [- \Delta,\Delta] \setminus \{ 0 \} $. As there is no danger of dividing by $ 0 $, we thus obtain
\begin{align}
\forall h \in [- \Delta,\Delta] \setminus \{ 0 \}: \qquad
~ & \frac{(f \circ g)(a + h) - (f \circ g)(a)}{g(a + h) - g(a)} \cdot
\frac{g(a + h) - g(a)}{h} \\
= ~ & \frac{(f \circ g)(a + h) - (f \circ g)(a)}{h}.
\end{align}
Equivalently,
\begin{align}
(\spadesuit) \qquad
\forall h \in [- \Delta,\Delta] \setminus \{ 0 \}: \qquad
~ & \frac{f(g(a + h)) - f(g(a))}{g(a + h) - g(a)} \\
= ~ & \frac{(f \circ g)(a + h) - (f \circ g)(a)}{h} \cdot
\frac{1}{\left[ \frac{g(a + h) - g(a)}{h} \right]}.
\end{align}
Define a function $ I: (0,\Delta] \to \mathcal{P}(\Bbb{R}) $ by
$$
\forall \delta \in (0,\Delta]: \quad
I(\delta) \stackrel{\text{df}}{=}
\{ g(a + h) \in \Bbb{R} \mid h \in [- \delta,\delta] \}.
$$
The Intermediate Value Theorem tells us that for each $ \delta \in (0,\Delta] $, the continuity of $ g $ guarantees that $ I(\delta) $ is a closed bounded interval, and as $ g(a + h) \neq g(a) $ for any $ h \in [- \delta,\delta] \setminus \{ 0 \} $, we see that $ I(\delta) $ is also non-degenerate, i.e, it contains points other than $ g(a) $. Next, define
\begin{align}
L & \stackrel{\text{df}}{=}
\{ \delta \in (0,\Delta] \mid I(\delta) \cap (- \infty,g(a)) \neq \varnothing \},
\\
R & \stackrel{\text{df}}{=}
\{ \delta \in (0,\Delta] \mid I(\delta) \cap (g(a),\infty) \neq \varnothing \}.
\end{align}
By the foregoing discussion, we have $ L \cup R = (0,\Delta] $. Hence, either
- $ 0 $ is a limit point of $ L $, or
- $ 0 $ is a limit point of $ R $.
Without any loss of generality, we may assume it is Case (1) that occurs.
Note: The cases are not mutually exclusive, i.e., both could occur.
By our assumption that Case (1) occurs, we have $ I(\Delta) \cap (- \infty,g(a)) \neq \varnothing $. As such, let $ (y_{n})_{n \in \Bbb{N}} $ be any sequence in $ I(\Delta) \cap (- \infty,g(a)) $ that converges to $ g(a) $. We boldly claim that
$$
\lim_{n \to \infty} \frac{f(y_{n}) - f(g(a))}{y_{n} - g(a)}
= \frac{(f \circ g)'(a)}{g'(a)},
$$
which would imply that the left-derivative of $ f $ at $ g(a) $ exists.
Define a sequence $ (h_{n})_{n \in \Bbb{N}} $ in $ [- \Delta,\Delta] $ by
$$
\forall n \in \Bbb{N}: \quad
h_{n} \stackrel{\text{df}}{=}
\text{A number $ h \in [- \Delta,\Delta] $ closest to $ 0 $ such that $ g(a + h) = y_{n} $}.
$$
Such a $ h $ exists because $ {g^{\leftarrow}}[\{ y_{n} \}] \cap [a - \Delta,a + \Delta] $ is a non-empty compact subset of $ \Bbb{R} $. So as to avoid using the Axiom of Choice at this stage, we choose $ h_{n} $ to be positive whenever possible.
We argue that $ \displaystyle \lim_{n \to \infty} h_{n} = 0 $. Assume the contrary. Then we can find an $ \epsilon > 0 $ and a subsequence $ (h_{n_{k}})_{k \in \Bbb{N}} $ of $ (h_{n})_{n \in \Bbb{N}} $ such that $ |h_{n_{k}}| \geq \epsilon $ for all $ k \in \Bbb{N} $. Choose a $ \delta \in (0,\epsilon) \cap L $ (this is where the assumption of Case (1) plays a role). As $ I(\delta) \cap (- \infty,g(a)) = [m,g(a)) $ for some $ m < g(a) $, we are able to find a $ K \in \Bbb{N} $ sufficiently large so that $ y_{n_{K}} \in I(\delta) $. It follows that $ y_{n_{K}} = g(a + h) $ for some $ h \in [- \delta,\delta] \subseteq (- \epsilon,\epsilon) $, which makes $ h $ even closer to $ 0 $ than $ h_{n_{K}} $ is — a contradiction.
Using $ (\spadesuit) $ now, we therefore get
\begin{align}
\lim_{n \to \infty} \frac{f(y_{n}) - f(g(a))}{y_{n} - g(a)}
& = \lim_{n \to \infty} \frac{f(g(a + h_{n})) - f(g(a))}{g(a + h_{n}) - g(a)} \\
& = \lim_{n \to \infty}
\frac{(f \circ g)(a + h_{n}) - (f \circ g)(a)}{h_{n}} \cdot
\frac{1}{\left[ \frac{g(a + h_{n}) - g(a)}{h_{n}} \right]} \\
& = (f \circ g)'(a) \cdot \frac{1}{g'(a)} \qquad
\left( \text{As $ \lim_{n \to \infty} h_{n} = 0 $.} \right) \\
& = \frac{(f \circ g)'(a)}{g'(a)}.
\end{align}
Therefore, $ f $ has a left-derivative at $ g(a) $. If we were to assume Case (2) instead, it would have a right-derivative at $ g(a) $. In any case, $ f $ has a one-sided derivative at $ g(a) $, and as we have shown in the beginning that there are uncountably many such $ a $, we are done. $ \quad \blacksquare $
