I personally prefer to work with sheaves which ultimately gives you the same thing, but sometimes you need to work in a dual setting. So, I will use sheaves below, feel free to rewrite this for yourself in the language of bundles.

Let $\alpha: \mathscr E_1\to \mathscr E_2$ be a surjective morphism of sheaves and a scheme $X$ and let $\pi_i:\mathbb P(\mathscr E_i)\to X$ be the associated projective bundle for $i=1,2$.

By construction there exist surjective sheaf morphisms $\pi_i^*\mathscr E_i\to \mathscr O_{\mathbb P(\mathscr E_i)}(1)$ on $\mathbb P(\mathscr E_i)$. Pulling back $\alpha$ to $\mathbb P(\mathscr E_2)$ gives a composition of surjective morphisms: $$ \pi_2^*\mathscr E_1\to \pi_2^*\mathscr E_2\to \mathscr O_{\mathbb P(\mathscr E_2)}(1). $$

This induces a natural map $\beta: \mathbb P(\mathscr E_2)\to \mathbb P (\mathscr E_1)$ such that $\beta^* \mathscr O_{\mathbb P(\mathscr E_2)}(1) = \mathscr O_{\mathbb P(\mathscr E_2)}(1)$. I leave it for you to check that $\beta$ is injective (after all you claimed this as a matter-of-fact in the question). The pull-back equality means that the tautological class of $\pi_1$ restricts to the tautological class of $\pi_2$.

In other words, using your notation, $\xi|_{\mathbb P(\mathscr E')}=\xi'$.

I personally prefer to work with sheaves which ultimately gives you the same thing, but sometimes you need to work in a dual setting. So, I will use sheaves below, feel free to rewrite this for yourself in the language of bundles.

Let $\alpha: \mathscr E_1\to \mathscr E_2$ be a surjective morphism of sheaves and a scheme $X$ and let $\pi_i:\mathbb P(\mathscr E_i)\to X$ be the associated projective bundle for $i=1,2$.

By construction there exist surjective sheaf morphisms $\pi_i^*\mathscr E_i\to \mathscr O_{\mathbb P(\mathscr E_i)}(1)$ on $\mathbb P(\mathscr E_i)$. Pulling back $\alpha$ to $\mathbb P(\mathscr E_2)$ gives a composition of surjective morphisms: $$ \pi_2^*\mathscr E_1\to \pi_2^*\mathscr E_2\to \mathscr O_{\mathbb P(\mathscr E_2)}(1). $$

This induces a natural map $\beta: \mathbb P(\mathscr E_2)\to \mathbb P (\mathscr E_1)$ such that $\beta^* \mathscr O_{\mathbb P(\mathscr E_2)}(1) = \mathscr O_{\mathbb P(\mathscr E_2)}(1)$. I leave it for you to check that $\beta$ is injective (after all you claimed this as a matter-of-fact in the question). The pull-back equality means that the tautological class of $\pi_1$ restricts to the tautological class of $\pi_2$.

In other words, using your notation, $\xi|_{\mathbb P(E')}=\xi'$.