Let $X,Y$ be real vector spaces, $T: X\to Y$ be a linear map, and fix a nonempty $S\subseteq X$ (we do not assume that $S$ is neither convex nor compact (indeed, right now we do not assume any topological structure on $X$ and $Y$)). Everything below is just basic linear algebra, hence you may prefer to go directly to the final question.
A point $x \in S$ is called an extreme point if $x=\alpha y+(1-\alpha)z$, for some $\alpha \in (0,1)$ and $y,z \in S$, implies $y=z$. Denote the set of extreme points with $\mathrm{Ext}_X(S)$. The following question is related to the commutation of $T$ and the $\mathrm{Ext}_X$ operator, cf. this and this one.
If $T$ is an injection then the condition $Tx=\alpha Ty+(1-\alpha)Tz$, with $x,y,z \in S$ and $\alpha \in (0,1)$, is equivalent to $x=\alpha y+(1-\alpha)z$, and similarly $Ty=Tz$ is equivalent to $y=z$. Hence $Tx$ is an extreme point of $T[S]$ if and only if $x$ an extreme point of $S$. Hence:
Fact. If $T \in \mathcal{L}(X,Y)$ is injective, then $\mathrm{Ext}_Y(T[S])=T[\mathrm{Ext}_X(S)]$.
Now, let us suppose that $T$ is not necessarily injective. Define $\hat{X}:=X/\mathrm{Ker}(T)$ be the quotient vector space and $\hat{T}: \hat{X}\to Y$ by $\hat{T}(x+\mathrm{Ker}(T))=Tx$ for all $x \in X$. Then $\hat{T}$ is a well-defined linear injection. We obtain by the Fact above (applied to $\hat{T}$ and $S+\mathrm{Ker}(T)$) that: $$ \mathrm{Ext}_{Y}(\hat{T}[S+\mathrm{Ker}(T)])=\hat{T}[\mathrm{Ext}_{\hat{X}}(S+\mathrm{Ker}(T))], $$ so that $$ \mathrm{Ext}_{Y}(T[S])=\hat{T}[\mathrm{Ext}_{\hat{X}}(S+\mathrm{Ker}(T))]. $$ (As it has been observed in the linked thread, the linear map $T:\mathbf{R}^2\to \mathbf{R}$ defined by $T(x,y)=x$ is such that $\mathrm{Ext}_Y(T[S])\subsetneq T[\mathrm{Ext}_X(S)]$ if $S$ is the closed unit ball of $\mathbf{R}^2$ with the $1$-norm. However, the above identity is correct since $\mathrm{Ext}_{Y}(T[S])=\mathrm{Ext}_T([-1,1])=\{-1,1\}$ and $\hat{T}[\mathrm{Ext}_{\hat{X}}(S+\mathrm{Ker}(T))]=\hat{T}(\mathrm{Ext}_{\hat{X}}([-1,1]\times \mathbf{R}))=\{-1,1\}$.)
Preliminary Question 1. Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\mathrm{Ext}_Y(T[S])\subseteq T[\mathrm{Ext}_X(S)]$?
Considering the above premises, this can be rewritten equivalently as follows:
Preliminary Question 2. Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\hat{T}[\mathrm{Ext}_{\hat{X}}(S+\mathrm{Ker}(T))]\subseteq T[\mathrm{Ext}_X(S)]$?
Suppose that the left hand side is nonempty (otherwise there is nothing to prove) and pick $x+\mathrm{Ker}(T)\in \mathrm{Ext}_{\hat{X}}(S+\mathrm{Ker}(T))$, i.e., if $\hat{T}(x+\ker{T})=\alpha \hat{T}(y+\mathrm{ker}(T))+(1-\alpha) \hat{T}(z+\mathrm{ker}(T))$, for some $\alpha \in (0,1)$, then $y-z \in \mathrm{Ker}(T)$. Equivalently, if $Tx=\alpha Ty+(1-\alpha)Tz$ then $Ty=Tz$. We would like to show, if the Question has a positive answer, then $Tx \in T[\mathrm{Ext}_X(S)]$, i.e., there exists $h \in \mathrm{Ext}_X(S)$ such that $Tx=Th$. However, this suggested the counterexample: pick $T:\mathbf{R}^2\to \mathbf{R}$ as the example above and $S:=\{0\}\times (0,1)$, so that $$ \mathrm{Ext}_Y(T[S])=\{0\}\not\subseteq \emptyset= T[\mathrm{Ext}_X(S)] $$ Hence, the Preliminary Questions have a negative answer. To avoid similar conterexamples with empty sets, we assume that $X,Y$ are Banach spaces, $T$ is continuous, $S$ (and so also $T(S)$) is compact convex. Hence we come to our final question:
Question Suppose that $X,Y$ are Banach spaces, $T$ is continuous, and $S\subseteq X$ is compact convex. Is it true that $\mathrm{Ext}_Y(T[S])\subseteq T[\mathrm{Ext}_X(S)]$?
For each $x \in S$, the coset $x+\mathrm{Ker}(T)$ is closed and convex, hence $H:=(x+\mathrm{Ker}(T)) \cap S$ is nonempty compact convex. Continuing the reasoning above, the answer should be positive in Hilbert spaces, picking the unique $h=h(x) \in H$ which minimizes the distance of $H$ from the origin (and, more generally, in Banach spaces with a strictly convex norm). Is it correct also in general Banach spaces?
