In order to show tightness, fix $\varepsilon>0$. Then you get $N=N(\varepsilon)$ such that $\int_Xf_nd\mu\leq\varepsilon$ if $n\geq N+1$. Now, for all $n\leq N$, you can find a positive $M$ such that $\int_{\{f_n\geq M_n\}}f_nd\mu\leq \varepsilon$, using integrability of $f_n$. (if $f$ is integrable apply the monotone convergence theorem to $f\mathbf 1_{\{|f|\geq n\}})$
Put $A_n:=\{f_n\leq M_n\}$, then $A_n$ is measurable. Take $X_0:=\bigcap_{k=1}^NA_k^c$. Each $A_k^c$ has finite measure (since $\mu(A_k^c)\leq \frac 1{M_k}\int f_kd\mu$) so $X_0$ is of finite measure. Check that we have the wanted inequality.
http://en.wikipedia.org/wiki/Uniform_integrability. Of course, there are some obvious conditions, like monotonicity of the sequence $\{f_i\}$. Here by monotonicity it is meant that for any measurable $A\subset X$, $\|f_n|_A\|_{L^1}\leq \|f_m|_A\|_{L^1}$ whenever $n\geq m$. Check that link for sufficient conditions.
The most general sufficient conditions are hard to come up with, simply because there are easy counterexamples even in the nicest cases (i.e. where you're dealing with a sequence of, say, continuous functions). As far as particular cases: there would be too many to list. Maybe you can tell us more about the sequence you have in mind?
EDIT: The answer above, and most of the discussion below, pertains to the general case before the OP posted the extra conditions (i.e. the stuff about function $g$), and clarified what is meant by local boundedness of $g$. Below is the answer pertaining to the specific family $f_i(\cdot) = g(z_i, \cdot)$.
Case i.
If $g$ is locally bounded independently of the second component, that is, if for every $z\in \mathbb{R}^n$ there exists $\delta > 0$ such that for every $w\in\mathbb{B}(z, \delta)$ and almost every $x\in X$, $|g(w,x)|\leq M$, then the answer to the original question is obviously yes. Indeed, then there exists $\delta_0 > 0$ and $M > 0$ such that for all $w\in\mathbb{B}(z, \delta_0)$ (with $z$ as in the original question), and almost every $x\in X$, $|g(w, x)|\leq M$. In particular, if $\{z_i\}$ is a sequence in $\mathbb{B}(z,\delta_0)$, then $|f_i|\leq M$ almost surely on $X$ (as the OP said, $\delta$ in the original post can be taken arbitrarily small, in particular we may take $\delta = \delta_0$). But then obviously we have uniform integrability.
Case ii.
Now suppose we're dealing with a less trivial example. Assume that $X$ is a topological space. Suppose that $g$ is such that for any $(z,x)\in\mathbb{R}^n\times X$, there exists an open neighborhood $\mathbb{B}$ around $(z, x)$ in $\mathbb{R}^n\times X$ and $M > 0$, such that on $\mathbb{B}$, $|g|\leq M$ (that is, local boundedness of $g$ depends on both components). Then the following counterexample illustrates that uniform integrability generally doesn't hold.
Define $g$ as follows. Let $X = \mathbb{R}_{> 0}$. Let $g(y,x) = f_x(y)$, where $f_x(y) = 0$ on $(0, 1/\|x\|]\cup [1/\|x\| + \|x\|^{1/2}, \infty)$, and $f_x(y) = 1/\|x\|$ on $(1/\|x\|, 1/\|x\| + \|x\|^{1/2})$.
Now take $\delta < 1$, and consider $g$ on $\mathbb{B}(0, \delta)$. Pick any $v\in\mathbb{R}^n$ with $\|v\| = 1$ and define the sequence $\{f_i\}_{i\in \mathbb{N}}$ by $f_i(x) = g(v/i, x)$.
Best Answer
Take $f_n = n 1_{A_n}$ where $\mu(A_n) = 1/n$. (For instance, on $[0,1]$ with Lebesgue measure, you could take $A_n = [0,\frac{1}{n}]$.) I think this is the example that was intended in the linked answer, and it is not uniformly integrable in your sense (or any other).