This problem can be solved by considering a time-inhomogeneous Markov chain. Let $t$ represent the time at which $t$ balls have been drawn, and define
\begin{align*}
X_t &= \text{number of distinct labels of blue balls drawn so far (i.e. by time $t$)}\\
Y_t &= \text{number of blue balls in first bag with labels not already drawn}\\
Z_t &= \text{number of blue balls in second bag with labels not already drawn}\\
\end{align*}
The initial state is $(X_0, Y_0, Z_0) = (0, n_1, n_2)$. We need to consider the transition probabilities at time $t$.
For even $t$, say $t=2s$, the next draw will be from the first bag. At this time, the first bag has $Y_t$ blue balls and $m-s$ total balls, so with probability $Y_t/(m-s)$ a blue ball will be drawn with a label not previously drawn. In this case, under the random labeling assumption, with probability $Z_t/(m-X_t)$ the label will match one of the blue balls in the second bag. So we can describe the transition probabilities for even $t$ as follows:
\begin{align*}
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x, y, z) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad= 1-\frac{Y_{t}}{m-s}\\ \\
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x+1, y-1, z) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad= \frac{Y_{t}}{m-s}\left(1-\frac{Z_{t}}{m-X_{t}}\right)\\ \\
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x+1, y-1, z-1) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad= \frac{Y_{t}}{m-s}\frac{Z_{t}}{m-X_{t}}
\end{align*}
For odd $t$, say $t=2s-1$, the next draw will be from the second bag, and we similarly have
\begin{align*}
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x, y, z) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad = 1-\frac{Z_{t}}{m-s}\\ \\
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x+1, y, z-1) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad = \frac{Z_{t}}{m-s}\left(1-\frac{Y_{t}}{m-X_{t}}\right)\\ \\
&P((X_{t+1}, Y_{t+1}, Z_{t+1}) = (x+1, y-1, z-1) \ |\ (X_{t}, Y_{t}, Z_{t}) = (x, y, z)) \\
&\quad = \frac{Z_{t}}{m-s}\frac{Y_{t}}{m-X_{t}}
\end{align*}
In this way, beginning with $t=0$ the pmf of $(X_{t+1}, Y_{t+1}, Z_{t+1})$ can be computed in terms of the pmf of $(X_t, Y_t, Z_t)$. Summing up over all possible values of $Y_t$ and $Z_t$, we obtain the (marginal) pmf of $X_t$. If we let $J$ denote the required number of draws, then $P(J \leq t) = P(X_t \geq k)$, so we may find the pmf of $J$ using the pmf of $X_t$. Here is some (not particularly efficient) R code for carrying this out:
prob = function(m, n1, n2){
p = array(0, dim = c(2*m+1, n1+n2+1, n1+1, n2+1))
p[1, 1, n1+1, n2+1] = 1
for(i in 0:(2*m-1)){
for(x in 0:(n1+n2)){
for(y in 0:n1){
for(z in 0:n2){
if(i%%2 == 0){
s = i/2
q = p[i+1, x+1, y+1, z+1]
if(q>0){
p[i+2, x+1, y+1, z+1] = p[i+2, x+1, y+1, z+1] + q * (1 - y/(m-s))
if(y>0){
p[i+2, x+2, y, z+1] = p[i+2, x+2, y, z+1] + q * y/(m-s) * (1 - z/(m-x))
p[i+2, x+2, y, z] = p[i+2, x+2, y, z] + q * y/(m-s) * z/(m-x)
}
}
}else{
s = (i-1)/2
q = p[i+1, x+1, y+1, z+1]
if(q>0){
p[i+2, x+1, y+1, z+1] = p[i+2, x+1, y+1, z+1] + q * (1 - z/(m-s))
if(z>0){
p[i+2, x+2, y+1, z] = p[i+2, x+2, y+1, z] + q * z/(m-s) * (1 - y/(m-x))
p[i+2, x+2, y, z] = p[i+2, x+2, y, z] + q * z/(m-s) * y/(m-x)
}
}
}
}
}
}
}
p
}
pmf_J = function(m, n1, n2, k){
p = prob(m, n1, n2)
cdf = numeric(2*m+1)
for(i in 0:(2*m)){
cdf[i+1] = sum(p[i+1, k:(n1+n2)+1,,])
}
diff(cdf)
}
And here are some example plots, using the same parameter combinations as in whuber's answer:
params = expand.grid(k = c(4, 12, 20), n1 = c(12, 19, 23), m = 50)
params$n2 = 24 - params$n1
png("charts.png", width=750, height=750)
par(mfrow = c(3,3), cex = 0.9)
for(i in 1:nrow(params)){
pa = params[i,]
pmf = pmf_J(pa$m, pa$n1, pa$n2, pa$k)
print(sum(pmf))
plot(1:(2*pa$m), pmf,
xlab = "j",
ylab = "Probability",
main = paste0("m=",pa$m,", n1=",pa$n1,", n2=",pa$n2,", k=",pa$k))
}
dev.off()

For parameter choices where at least one of $n_1$, $n_2$, or $k$ is small compared to $m$, there is little impact from the condition that a blue ball with a repeated label does not count, and so the plots agree fairly closely in these cases. The starkest difference can be seen in the upper-right chart ($n_1= n_2 = 12$, $k=20$). Here $k$ exceeds $\max\{n_1, n_2\}$ (contrary to the assumption in the question), which means that there is positive probability that all balls in both bags will be exhausted without $k$ distinctly labeled blue balls being drawn; in this case, we could set $J = \infty$, but the chart only shows the probabilities for finite $J$.