The answer is yes, in full generality, after possibly restricting to a smaller open set where $P$ admits a principal bundle chart. The restriction to a smaller open set can be seen to be essential by considering the case where $P'$ is the (unique) trivial-group bundle over $M$.

This is straightforward to see using the perspective that the existence of a section $\sigma$ of a principal bundle $P$ on an open set $U$ is equivalent to $P$ admitting a principal bundle chart on $U$. (Given a section $\sigma$ for $x \in U$, every point in the fiber over $x$ can be written uniquely as $\sigma(x)g$ for some $g \in G$. The chart is $\sigma(x)g \mapsto (x,g)$. Conversely, take the identity section of the product chart).

So let $U \subset M$ be a neighborhood so $P$ has a local section $\sigma$ on $U$. Then, in your notation, $F \circ \sigma$ is a section of $P'$ on $U$, and the $f$-compatibility condition $F(\sigma(x)g) = F(\sigma(x)) f(g)$ gives exactly what you want in the associated charts for $\sigma$ and $F \circ \sigma$ as above.

EDIT: If you want a chart for $P$ compatible with the specific chart $\phi$ on $U$, this can be arranged for surjective homomorphisms $f$ by using the fact that Lie group homomorphisms have constant rank, and so surjections admit local models based on standard projections $\mathbb{R}^n \to \mathbb{R}^k$ in a neighborhood of the $e$. In particular, if we denote by $V_G, V_H$ neighborhoods of $e$ in $G,H$ so $f|_{V_G}$ has a standard model in charts for $V_G, V_H$, there is a smooth section $\nu: V_H \to V_G$ of $f$, i.e. one has $f \circ \nu = \text{id}$.

To use this, let $\sigma'$ be the section corresponding to $\phi$. Given any two sections $\sigma_1, \sigma_2$ of $P'$ on $U$, they differ by a map $\alpha: U \to H$ (i.e. so $\sigma_1(x) = \sigma_2(x)\alpha(x))$. So let $\alpha$ be so $\sigma'(x) = F(\sigma(x))\alpha(x)$ for $x \in U$. Pick $p \in U$. Modifying $\sigma$ by some $g \in G$, we can arrange for $F(\sigma(p)) = \sigma'(p)$. In a neighborhood $U''$ of $p$, in the chart for $P'$ induced by $F \circ \sigma$, the section $\sigma'$ is valued in $U \times V_H$, and so $\nu: V_H \to V_G$ induces a section $\eta: U \to P$ by specifying $\eta(x) = \sigma(x)\nu(\alpha(x))$ for $x \in U''$. Then use $\eta$ in place of $\sigma$ in the above construction to get the desired chart. By construction, $F(\eta(x)) = F(\sigma(x))f(\nu(\alpha(x))) = \sigma'(x)$, so one obtains a compatible chart for $P$.

The answer is yes, in full generality, after possibly restricting to a smaller open set where $P$ admits a principal bundle chart. The restriction to a smaller open set can be seen to be essential by considering the case where $P'$ is the (unique) trivial-group bundle over $M$.

This is straightforward to see using the perspective that the existence of a section $\sigma$ of a principal bundle $P$ on an open set $U$ is equivalent to $P$ admitting a principal bundle chart on $U$. (Given a section $\sigma$ for $x \in U$, every point in the fiber over $x$ can be written uniquely as $\sigma(x)g$ for some $g \in G$. The chart is $\sigma(x)g \mapsto (x,g)$. Conversely, take the identity section of the product chart).

So let $U \subset M$ be a neighborhood so $P$ has a local section $\sigma$ on $U$. Then, in your notation, $F \circ \sigma$ is a section of $P'$ on $U$, and the $f$-compatibility condition $F(\sigma(x)g) = F(\sigma(x)) f(g)$ gives exactly what you want in the associated charts for $\sigma$ and $F \circ \sigma$ as above.

EDIT: If you want a chart for $P$ compatible with the specific chart $\phi$ on $U$, this can be arranged for surjective homomorphisms $f$ by using the fact that Lie group homomorphisms have constant rank, and so surjections admit local models based on standard projections $\mathbb{R}^n \to \mathbb{R}^k$ in a neighborhood of $e$. In particular, if we denote by $V_G, V_H$ neighborhoods of $e$ in $G,H$ so $f|_{V_G}$ has a standard model in charts for $V_G, V_H$, there is a smooth section $\nu: V_H \to V_G$ of $f$, i.e. one has $f \circ \nu = \text{id}$.

To use this, let $\sigma'$ be the section corresponding to $\phi$. Given any two sections $\sigma_1, \sigma_2$ of $P'$ on $U$, they differ by a map $\alpha: U \to H$ (i.e. so $\sigma_1(x) = \sigma_2(x)\alpha(x))$. So let $\alpha$ be so $\sigma'(x) = F(\sigma(x))\alpha(x)$ for $x \in U$. Pick $p \in U$. Modifying $\sigma$ by some $g \in G$, we can arrange for $F(\sigma(p)) = \sigma'(p)$. In a neighborhood $U''$ of $p$, in the chart for $P'$ induced by $F \circ \sigma$, the section $\sigma'$ is valued in $U \times V_H$, and so $\nu: V_H \to V_G$ induces a section $\eta: U \to P$ by specifying $\eta(x) = \sigma(x)\nu(\alpha(x))$ for $x \in U''$. Then use $\eta$ in place of $\sigma$ in the above construction to get the desired chart. By construction, $F(\eta(x)) = F(\sigma(x))f(\nu(\alpha(x))) = \sigma'(x)$, so one obtains a compatible chart for $P$.

The answer is yes, in full generality, after possibly restricting to a smaller open set where $P$ admits a principal bundle chart.

This is straightforward to see using the perspective that the existence of a section $\sigma$ of a principal bundle $P$ on an open set $U$ is equivalent to $P$ admitting a principal bundle chart on $U$. (Given a section $\sigma$ for $x \in U$, every point in the fiber over $x$ can be written uniquely as $\sigma(x)g$ for some $g \in G$. The chart is $\sigma(x)g \mapsto (x,g)$. Conversely, products have sections).

So let $U \subset M$ be a neighborhood so $P$ has a local section $\sigma$ on $U$. Then, in your notation, $F \circ \sigma$ is a section of $P'$ on $U$, and the $f$-compatibility condition $F(\sigma(x)g) = F(\sigma(x)) f(g)$ gives exactly what you want in the associated charts for $\sigma$ and $F \circ \sigma$ as above.

The answer is yes, in full generality, after possibly restricting to a smaller open set where $P$ admits a principal bundle chart. The restriction to a smaller open set can be seen to be essential by considering the case where $P'$ is the (unique) trivial-group bundle over $M$.

This is straightforward to see using the perspective that the existence of a section $\sigma$ of a principal bundle $P$ on an open set $U$ is equivalent to $P$ admitting a principal bundle chart on $U$. (Given a section $\sigma$ for $x \in U$, every point in the fiber over $x$ can be written uniquely as $\sigma(x)g$ for some $g \in G$. The chart is $\sigma(x)g \mapsto (x,g)$. Conversely, take the identity section of the product chart).

So let $U \subset M$ be a neighborhood so $P$ has a local section $\sigma$ on $U$. Then, in your notation, $F \circ \sigma$ is a section of $P'$ on $U$, and the $f$-compatibility condition $F(\sigma(x)g) = F(\sigma(x)) f(g)$ gives exactly what you want in the associated charts for $\sigma$ and $F \circ \sigma$ as above.

EDIT: If you want a chart for $P$ compatible with the specific chart $\phi$ on $U$, this can be arranged for surjective homomorphisms $f$ by using the fact that Lie group homomorphisms have constant rank, and so surjections admit local models based on standard projections $\mathbb{R}^n \to \mathbb{R}^k$ in a neighborhood of the $e$. In particular, if we denote by $V_G, V_H$ neighborhoods of $e$ in $G,H$ so $f|_{V_G}$ has a standard model in charts for $V_G, V_H$, there is a smooth section $\nu: V_H \to V_G$ of $f$, i.e. one has $f \circ \nu = \text{id}$.

To use this, let $\sigma'$ be the section corresponding to $\phi$. Given any two sections $\sigma_1, \sigma_2$ of $P'$ on $U$, they differ by a map $\alpha: U \to H$ (i.e. so $\sigma_1(x) = \sigma_2(x)\alpha(x))$. So let $\alpha$ be so $\sigma'(x) = F(\sigma(x))\alpha(x)$ for $x \in U$. Pick $p \in U$. Modifying $\sigma$ by some $g \in G$, we can arrange for $F(\sigma(p)) = \sigma'(p)$. In a neighborhood $U''$ of $p$, in the chart for $P'$ induced by $F \circ \sigma$, the section $\sigma'$ is valued in $U \times V_H$, and so $\nu: V_H \to V_G$ induces a section $\eta: U \to P$ by specifying $\eta(x) = \sigma(x)\nu(\alpha(x))$ for $x \in U''$. Then use $\eta$ in place of $\sigma$ in the above construction to get the desired chart. By construction, $F(\eta(x)) = F(\sigma(x))f(\nu(\alpha(x))) = \sigma'(x)$, so one obtains a compatible chart for $P$.