Here are the definitions of the extension and trace of a filter (paraphrased from IM James' book "Topological and Uniform Structures"):
Let $\mathcal{F}$ be a filter on the set $X$. For each set $X'$ and function $\phi: X \to X'$ a filter $\phi_*\mathcal{F}$ on $X'$ is defined by taking the direct images of the members of $\mathcal{F}$ as a filter base. In case $X'$ is a superset of $X$ and $\phi$ the inclusion we refer to $\phi_*\mathcal{F}$ as the extension of $\mathcal{F}$ to $X'$.
Let $\mathcal{F}$ be a filter on the set $X'$. For each set $X$ and each function $\phi: X' \to X$, if $\phi^{-1}(A) \neq \emptyset$ for every $A \in \mathcal{F}$, then a filter $\phi^*\mathcal{F}$ on $X$ is defined by taking the inverse images of the members of $\mathcal{F}$ as a filter base. In case $X'$ is a subset of $X$ and $\phi$ the inclusion we refer to $\phi^*\mathcal{F}$ as the trace of $\mathcal{F}$ on $X'$.
Consider the inclusion map $\iota: \mathbb{R} \hookrightarrow\mathbb{R}^2$ with the standard topologies and let $\mathcal{F}$ be the neighbourhood filter of the origin in $\mathbb{R}$. The open unit ball $B$ in $\mathbb{R}^2$ is not a direct image of any $A \in \mathcal{F}$, but $\phi((0,1)) \subseteq B$ implies that $B$ is in $\phi_*\mathcal{F}$.
Thus in general $\phi_*\mathcal{F} \neq \phi(\mathcal{F})$ i.e. extensions are not the same as direct images.
Problem: What about the preimage of a filter $\phi^{-1}(\mathcal{F}) = \{\, A \mid \phi(A) \in \mathcal{F}\,\}$ and the trace $\phi^*\mathcal{F}$?
We certainly have $\phi^*\mathcal{F} \supseteq \phi^{-1}(\mathcal{F})$. But in general, does equality hold?
No, a counterexample (almost your example): Let $\phi$ be the inclusion map from $X=\mathbb{R} \times \{0\}$ into $X'=\mathbb{R}^2$, and let $\mathcal{F}$ be the neighbourhood filter of $(0,0)$ on $X'$ (usual topology, and $X$ gets the subspace topology).
Then $(-1,1) \times \{0\}$ is in $\phi^\ast \mathcal{F}$ because it equals $B((0,0),1) \cap X = \phi^{-1}[B((0,0),1)]$, (where $B(x,r)$ denotes the metric balls), but it is not in $\phi^{-1}[\mathcal{F}]$ because $\phi[(-1,1) \times \{0\}] = (-1,1) \times \{0\} \notin \mathcal{F}$ (it contains no ball around the origin). If you think about it, the way you have defined it, $\phi^{-1}[\mathcal{F}] = \emptyset$, so that type of inverse image (preimage) is not really a useful one for filters. The pullback (as $\phi^\ast \mathcal{F}$ is usually called) is more natural. For inclusion maps it comes down to the intersections of filter members with a subset, hence your name "trace": if $\mathcal{P}$ is a family of subsets of $X$ and $A \subseteq X$ the trace of $\mathcal{P}$ on $A$ is by definition $\{P \cap A: P \in \mathcal{P}\}$. For filters we then have the condition that all these intersection must be non-empty, to avoid the trivial filter.