Is the "evaluation map" $Y^{I}\times I\to Y$ continuous for general $Y$?

Yes. It is only needed that $I$ is strongly (see the comment of Henno) locally compact.

Let $e$ denote the evaluation function, let $U$ be an open subset of $Y$ and let $e(\gamma,t)=\gamma(t)\in U$.

Since $\gamma$ is continuous and $t$ has arbitrarily small compact neighborhoods a compact neighborhood $K$ of $t$ can be found in $I$ with $\gamma(K)\subseteq U$ or equivalently $\gamma\in M(K,U)$ where $M(K,U)$ is defined as the set $\{f\in Y^I\mid f(K)\subseteq U\}$ and belongs to the subbase of the compact-open topology on $Y^I$.

Then $M(K,U)\times K$ is a neighborhood of $\langle\gamma,t\rangle$ in $Y^I\times I$ that satisfies $e(M(K,U)\times K)\subseteq U$.

This proves that $e$ is continuous at arbitrary $\langle\gamma,t\rangle\in Y^I\times I$, so we conclude that $e$ is continuous.

It was not used in this proof that $Y$ is Hausdorff.