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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. (It seems likely that, if we get a non-trivial multiple as the weight, then we can just extract a root; but I do not yet see why this is so.) Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation.

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$. Computing in $\SL_2$ shows that there are a representative $n_j$ in $N_G(T)(k)$ of $s_j$, and functions $u_\ell, u_r : \GL_1 \to U_j$, such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t))$ is non-$0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_j(1)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have that $\langle\chi, \alpha_j^\vee\rangle$ is non-negative; and $\langle\chi, \alpha_j^\vee\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. (It seems likely that, if $\chi$ is a non-trivial multiple $N\chi_i$ of $\chi_i$, then a function $F$ on $B w_0 B$ with weight $(-w_0\chi_i, \chi_i)$ such that $F(n_0)^N$ equals $f(n_0)$ extends to $G$; but I do not yet see for sure why this is so.) Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation.

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$. Computing in $\SL_2$ shows that there are a representative $n_j$ in $N_G(T)(k)$ of $s_j$, and functions $u_\ell, u_r : \GL_1 \to U_j$, such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$.

Per the comments, here are the deails. We have that $\operatorname C_G\bigl(\ker(\alpha_j)^\circ_\text{smooth}\bigr)$ is a Levi subgroup of $G$, so its derived subgroup $G_j$ is a rank-$1$, simply connected group, hence is $\SL_2$. We may choose the isomorphism so that it carries $\alpha_j^\vee(t)$ to $\operatorname{diag}(t, t^{-1})$, and $U_j$ to $\left\{\begin{pmatrix} 1 & * \\ & 1 \end{pmatrix}\right\}$. Write $n_j$, $u_\ell(t)$, and $u_r(t)$ for the pre-images in $G$ of $\begin{pmatrix} & 1 \\ -1 \end{pmatrix}$, $\begin{pmatrix} 1 & (1 + t)t^{-1} \\ & 1 \end{pmatrix}$, and $\begin{pmatrix} 1 & (1 - t)t^{-1} \\ & 1 \end{pmatrix}$, respectively. Then, if I have not messed things up, we have that $u_\ell(t)n_j\alpha_j^\vee(t)u_r(t)$ is the pre-image of $\begin{pmatrix} 1 + t & -t \\ t & 1 - t \end{pmatrix}$.

With these definitions $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t))$ is non-$0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_j(1)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have that $\langle\chi, \alpha_j^\vee\rangle$ is non-negative; and $\langle\chi, \alpha_j^\vee\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation. (Actually I am a little confused overall, because it seems to me that, if $f$ defines the closure of $B w_0 s_i B$, then so does any positive integer power of $f$. But, if I am not just saying something absurd, probably this is fixed by some implicit condition like taking $f$ to be a generator of the appropriate ideal—maybe after all that is what "defined by" means.)

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$, and choose a representative $n_j$ for $s_j$ in $N_G(T)(k)$. Computing in $\SL_2$ shows that there are functions $u_\ell, u_r : \GL_1 \to U_j$ such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t)) \ne 0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_1(t)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have $\langle\chi, \alpha_j\rangle \ge 0$. We have that $\langle\chi, \alpha_j\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. (It seems likely that, if we get a non-trivial multiple as the weight, then we can just extract a root; but I do not yet see why this is so.) Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation.

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$. Computing in $\SL_2$ shows that there are a representative $n_j$ in $N_G(T)(k)$ of $s_j$, and functions $u_\ell, u_r : \GL_1 \to U_j$, such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t))$ is non-$0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_j(1)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have that $\langle\chi, \alpha_j^\vee\rangle$ is non-negative; and $\langle\chi, \alpha_j^\vee\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation.

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$, and choose a representative $n_j$ for $s_j$ in $N_G(T)(k)$. Computing in $\SL_2$ shows that there are functions $u_\ell, u_r : \GL_1 \to U_j$ such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t)) \ne 0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_1(t)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have $\langle\chi, \alpha_j\rangle \ge 0$. We have that $\langle\chi, \alpha_j\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.

I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation. (Actually I am a little confused overall, because it seems to me that, if $f$ defines the closure of $B w_0 s_i B$, then so does any positive integer power of $f$. But, if I am not just saying something absurd, probably this is fixed by some implicit condition like taking $f$ to be a generator of the appropriate ideal—maybe after all that is what "defined by" means.)

Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.

For each $j$, write $U_j$ for the root group associated to $\alpha_j$, and choose a representative $n_j$ for $s_j$ in $N_G(T)(k)$. Computing in $\SL_2$ shows that there are functions $u_\ell, u_r : \GL_1 \to U_j$ such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t)) \ne 0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.

We have that $$ g_j(t) = U_\ell(t)^{-1}g_1(t)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have $\langle\chi, \alpha_j\rangle \ge 0$. We have that $\langle\chi, \alpha_j\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.