Show that $\forall x\varphi\vDash\varphi[t/x]$ may not hold if $t$ is bound for $x$ in $\varphi$.
Solution: Let $x=x_0$, $t=x_1$ and $\varphi=P(x_0,x_1)$. Then we have $\forall x_0(\exists x_1(P(x_0,x_1)))\vDash\exists x_1(P(x_0,x_1))[x_1/x_0]=\exists x_1 P(x_1,x_1)$. If we interpret the domain as the natural numbers and P as "less than" is this a valid counterexample? Because there clearly doesn't exists any natural numbers with the property that $x_1<x_1$. Let $x_1=x_0+1$ and so the left hand side is always true, that is, it isn't a logical consequence. Is this a valid counterxample? Can I let $\varphi$ be a formula which includes $P$. It's sometimes pretty hard for me to know what $\varphi$ really is. What it's allowed and not allowed to be. Thanks :)
What is meant by "for $x$" in "bound for $x$"? I guess it's just superfluous. Then you do have a valid counterexample, with the minor correction that you forgot the quantifier when giving $\varphi$; $\varphi$ actually is $\exists x_1 P(x_0, x_1)$ in your example.
$\varphi$ here is a metavariable, just as is $x$. This means that $\forall x \varphi$ isn't by itself a formula, but you can construct a formula by replacing $\varphi$ by any formula and $x$ by any quantifier variable you like. $\varphi$, therefore, may be as complex as you like, and may include $P$ if your language contains $P$.
Depending on how formal your justification of the counterexample should be, it may be perfectly fine to just say "interpret $P$ as the less-than predicate", and so on. But you shouldn't write "Let $x_1 = x_0 + 1$", or that would seem formally incorrect to me. The reason is that $x_1$ and $x_0$ are not constants that you can just assign values, but variables bound by quantifiers. Instead, as a (semi-formal) justification, you could for example say that any $x_0$ is less than its successor $x_0 + 1$, so for any $x_0$ there is an $x_1$ such that $P(x_0,x_1)$.
Doing it a bit more formal could look like this. Also, I here give an easier interpretation. As the domain of discourse, choose $D = \{1,2\}$. Now interpret $P$ such that $P(x_0, x_1) \Leftrightarrow x_0 \neq x_1$, that is:
$$P(1,1) = F \\ P(1,2) = T \\ P(2,1) = T \\ P(2,2) = F \\$$
Then, for all $x_0 \in D$, there is a $x_1 \in D$ such that $P(x_0, x_1)$, because $P(1,2)$ and $P(2,1)$ are both true. But it is not the case that there is a $x_1 \in D$ such that $P(x_1, x_1)$, because $P(1,1)$ and $P(2,2)$ are both false. So that's an interpretation which makes the premise true and the conclusion false, and we have a valid counterexample.