You can take $y_0 = 0$, $y_1 = -tx_1^m$, $y_2 = x_0^m$.
EDIT. Here is a simple computation of the normal bundle of the curve $$ C_t = \{([x_0:x_1],[0:-tx_1^m:x_0^m])\} \subset \mathbb{P}^1 \times \mathbb{P}^2 $$ in the surface $$ S_t = \{x_0^ny_1−x_1^ny_0+tx_0^{n−m}x_1^my_2=0\} $$ for $t \ne 0$. First, the normal bundle fits into the exact sequence $$ 0 \to N_{C_t/S_t} \to N_{C_t/\mathbb{P}^1 \times \mathbb{P}^2} \to N_{S_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert_{C_t} \to 0. $$ Since $C_t$ is a graph of a map $\mathbb{P}^1 \to \mathbb{P}^2$ (of degree $m$), its normal bundle is the pullback of the tangent bundle of $\mathbb{P}^2$, hence its degree is $$ \deg(N_{C_t/\mathbb{P}^1 \times \mathbb{P}^2}) = 3m. $$ On the other hand, $S_t$ is a divisor of type $(n,1)$ on $\mathbb{P}^1 \times \mathbb{P}^2$, hence its normal bundle restricted to $C_t$ has the degree equation which has degree $$ \deg(N_{S_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert_{C_t}) = n + m. $$ Therefore, $$ \deg(N_{C_t/S_t}) = 3m - (n + m) = 2m - n. $$