Log-linearization of the market clearing condition
We have:

$$y_t = c_t − w^u(1 − N_t) + \kappa v_t$$

We take the log of both sides:

$$\ln y_t = \ln \left[ c_t − w^u(1 − N_t) + \kappa v_t \right]$$

And then linearize around the steady states:

$$\begin{align}
\ln \bar{y} + \frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right] + \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$

Cancel $\ln \bar{y}$ and $\ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]$

$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$

Multiply the first term on the right hand side by $\frac{\bar{c}}{\bar{c}}$, and similarly for the other terms:

$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{\bar{c}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(c_t - \bar{c})}{\bar{c}} \\
&amp; + \frac{w_u \bar{N}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(N_t - \bar{N})}{\bar{N}} \\
&amp; + \frac{\kappa \bar{V}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(V_t - \bar{V})}{\bar{V}}
\end{align}
$$

And we simplify, taking advantage of the fact that $\hat{v}_t = \hat{\theta}_t + \hat{u}_t$, to get what you have:

$$\hat{y}_t = \frac{\bar{C}}{\bar{Y}}\hat{c}_t + w^u \frac{\bar{N}}{\bar{Y}} \hat{n}_t + \left( \frac{K\bar{V}}{\bar{Y}} \right) (\hat{\theta}_t + \hat{u}_t) $$

I don't think this derivation is wrong. But you can see that $y_t$ and $N_t$ are directly proportional to each other. It's definitely plausible that $\frac{\bar{N}}{\bar{Y}} = 1$, but I've pored over the equations in the setup for 3 hours to no avail at showing that rigorously. It's probably something silly that I don't understand. I'll look back later and if nothing else, I'll turn this answer into a community wiki so anyone who has the right idea can edit it.