As for your other question, I will work with complex Lie groups, but you can pass to maximal compact subgroups if you prefer.

Consider the natural embedding $\operatorname{SO}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{SO}_4$ is taken with respect to the quadratic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 + x_2 x_3$. If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the common restriction of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{Sp}_4(\mathbb C)$ is a simple (with respect to the upper-triangular Borel) root $a$, and the common restriction of $\alpha_1 + \alpha_2$ and $\alpha_2 + \alpha_3$ is a simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, $b^\vee$ equals $\alpha_1^\vee + 2\alpha_2^\vee + \alpha_3^\vee$, and $\Lambda$ equals $\mathbb Z a^\vee + \mathbb Z b^\vee$, but $(\Lambda' \cap \mathfrak t)/\Lambda$ has order $2$, generated by the image of $\alpha_2^\vee$.

As for your other question, I will work with complex Lie groups, but you can pass to maximal compact subgroups if you prefer.

Consider the natural embedding $\operatorname{SO}_4(\mathbb C) \to \operatorname{SL}_4(\mathbb C)$, where $\operatorname{SO}_4$ is taken with respect to the quadratic form $(x_1, x_2, x_3, x_4) \mapsto x_1 x_4 + x_2 x_3$. If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}_4(\mathbb C)$ by $\alpha_1, \alpha_2, \alpha_3$, in the obvious fashion, then the common restriction of $\alpha_1$ and $\alpha_3$ to the diagonal torus in $\operatorname{SO}_4(\mathbb C)$ is a simple (with respect to the upper-triangular Borel) root $a$, and the common restriction of $\alpha_1 + \alpha_2$ and $\alpha_2 + \alpha_3$ is a simple root $b$. Then $a^\vee$ equals $\alpha_1^\vee + \alpha_3^\vee$, $b^\vee$ equals $\alpha_1^\vee + 2\alpha_2^\vee + \alpha_3^\vee$, and $\Lambda$ equals $\mathbb Z a^\vee + \mathbb Z b^\vee$, but $(\Lambda' \cap \mathfrak t)/\Lambda$ has order $2$, generated by the image of $\alpha_2^\vee$.