I have determined
$$A=\{(X,Y,Z,T)\in V\mid Y\neq 0 \textrm{ or } T\neq 0\}\subset \textrm{dom }f$$
However I am unsure of how to show that $A=\textrm{dom }f$.
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Let $W = V(Y,T)$ and note that $W$ is contained in $V$. You showed that $V - W$ is contained in $\text{dom}\,f$. Here is how to see equality. The coordinates of $\mathbb{A}^4$ are $X$, $Y$, $Z$, $T$. Suppose that$$f = X/Y = Z/T$$in $k(V)$ is defined at a point $p = (a,0,b,0)$. Then there are $g$, $h$ in $k[X,Y,Z,T]$ such that $f = g/h$ in $k(V)$ and $h(p)$ is nonzero, so that $g(p)/h(p)$ is defined. Consider the following definitions.
Definition 1. Let $V$ be an irreducible affine variety in $k^n$. We call $QF(k[V])$ the function field (or field of rational functions) on $V$, and we denote this field by $k(V)$.
Definition 2. Let $V \subset k^m$ and $W \subset k^n$ be irreducible affine varieties. A rational mapping from $V$ to $W$ is a function $\phi$ represented by$$\phi(x_1, \ldots, x_m) = \left( {{f_1(x_2, \ldots, x_m)}\over{g_1(x_1, \ldots, x_m)}}, \ldots, {{f_n(x_1, \ldots, x_m)}\over{g_n(x_1, \ldots, x_m)}} \right),$$where $f_i/g_i \in k(x_1, \ldots, x_m)$ satisfy:
(i) $\phi$ is defined at some point of $V$.
(ii) For every $(a_1, \ldots, a_m) \in V$ where $\phi$ is defined, $\phi(a_1, \ldots, a_m) \in W$.
Definition 3. Let $\phi$, $\psi: V \mathrel{-\,}\rightarrow W$ be rational mappings represented by$$\phi = \left({{f_1}\over{g_1}}, \ldots, {{f_n}\over{g_n}}\right) \text{ and } \psi = \left({{h_1}\over{k_1}}, \ldots, {{h_n}\over{k_n}}\right).$$Then we say that $\phi = \psi$ if for each $i$, $1 \le i \le n$,$$f_ik_i - h_ig_i \in I(V).$$
Let us apply Definition 3 to $X/Y$ and $g/h$ and also to $Z/T$ and $g/h$. This gives $Xh -gY$, $Zh-gT \in \langle XT-YZ\rangle$, so$$Xh -gY = B(XT-YZ) \text{ and }Zh - gT = C(XT-YZ) \text{ for some polynomials }B,C \text{ in }k[X,Y,Z,T].$$Multiply the first equation by $T$ and the second by $Y$ and subtract. This gives $$h(XT-YZ) = (BT-CY)((XT-YZ) \implies h = BT-CY.$$Then $$h(p) = B(a,0,b,0) 0 - C(a,0,b,0) 0 = 0,$$which is a contradiction. A similar argument shows that $g = BZ-CX$, so that$$f = (BZ-CX)/(BT-CY).$$The geometric intuition is that on $V$, the only ways to represent the rational function $f$ are $(BZ-CX)/(BT-CY)$ (note that $X/Y$ and $Z/T$ are special cases of this). The set $W$ is where all of these representations fail to be defined.
