What is important here is that the $p$-values $p_1,\ldots,p_n$ are independent and constructed so that if the null $H_i$ is true, $p_i\sim U[0,1]$. They are not restricted to one-sided tests/p-values.
For example, for Fisher's test with test statistic
$$F=-2\sum_i\ln(p_i),$$
we obtain its null distribution ($\chi^2_{2n}$) as follows: let
$y=g(p_i):=-2\ln(p_i)$. Then, $p_i=g^{-1}(y)=e^{-\frac{1}{2}y}$
and the density of $-2\ln(p_i)$ is given by
$$ f_{-2\ln(p_i)}(y)=f_{p_i}(g^{-1}(y))\left|\frac{\partial}{\partial y}g^{-1}(y)\right|.
$$
Note
$$\frac{\partial}{\partial
y}g^{-1}(y)=-\frac{1}{2}e^{-\frac{1}{2}y}$$
and
$$\left|\frac{\partial}{\partial y}g^{-1}(y)\right|=\frac{1}{2}e^{-\frac{1}{2}y}.$$
We have $f_{p_i}(g^{-1}(y))=1\;\forall\;g^{-1}(y)\in[0,1]$ (a standard uniform density). This implies
$$f_{-2\ln(p_i)}(y)=\frac{1}{2}e^{-\frac{1}{2}y}.$$
The density of a $\chi^2_R$ random variable is
$$f_{\chi^2_R}(y)=\frac{1}{2^{R/2}\Gamma(R/2)}y^{\frac{R}{2}-1}e^{-\frac{y}{2}}.$$
With $R=2$, we get $f_{\chi^2_2}(y)=\frac{1}{2\Gamma(1)}e^{-\frac{y}{2}}.$ Recall that $\Gamma(1)=\int_0^\infty t^{1-1}e^{-t}\;dt=1$. So,
$$
f_{\chi^2_2}(y)=\frac{1}{2}e^{-\frac{y}{2}}.
$$
We have shown that $f_{-2\ln(p_i)}(y)=f_{\chi^2_2}(y)$. The proof is
complete since the sum of $n$ independent
$\chi^2_R$ r.v.s is distributed as $\chi^2_{n\cdot R}$.
That and why $p$-values are uniform under the null is for example discussed here: Why are p-values uniformly distributed under the null hypothesis?
My conjecture for your mistaken claim that you may only use one-sided p-values might be addressed here: Fisher's method for combing p-values - what about the lower tail?
The link explains that only large values of the test statistic provides evidence against the meta-null that all nulls $H_i$ are true, because the way the statistic is constructed, small $p_i$ translate into a large test statistic $F$.
