Prove that a continuous $\succsim$ is quasilinear
First off it seems to me you are proceeding in a manner more complicated than necessary. (Perhaps this is intentional because you wish to face a harder exercise.) Since it is given that $v(x)$ represents $\succsim$ you only have to show
$$
\forall x,x' \in X: v(x') \geq v(x) \iff u(x') \geq u(x).
$$
I think this would be a lot simpler than what you are currently doing.

On to your question:
 The function $u(x)$ is very well defined, as
$$
u(y,x_{L}) =  x_{L} + \phi(y) = x_{L} - x_L(y) 
$$
and we know that $v(y,x_L(y)) = 0$ describes the indifference curves of $\succsim$. The conditions you describe would violate this assumption because from
$$
(y,x_{L}) \precsim (y',x_{L}')
$$
it follows that
$$
v(y,x_{L}) \leq v(y',x_{L}').
$$
We can assume $v(y,x_{L}) = 0$, this is an affine transformation of any $v(y,x_{L})$. So $x_L(y) = x_{L}$.
 (I am not crazy about the notation, feel free to edit my post as long as it remains consistent with the question.)

 Then we also have $0 \leq v(y',x_{L}')$, and because of monotonicity this means
$x_L(y') &lt; x_{L}'$.
 Then
$$
u(y',x_{L}') &lt; u(y,x_{L})
$$
cannot hold as
$$
u(y,x_{L}) = x_{L} + \phi(y) = x_{L} - x_{L}(y) = 0
$$
and
$$
u(y',x_{L}') = x_{L}' + \phi(y') = x_{L}' - x_{L}(y') &gt; 0,
$$
because of what we have shown earlier, $x_L(y') &lt; x_{L}'$.