Is there a simple reason why the expected number of coin flips till getting $m$ more heads than tails or $n$ more tails than heads should be $mn$?

I'm not sure if this is a straightforward "intuitive reason" like you're hoping for, but it is a very different solution to this problem that is recursion-free and (I think) quite fun. Let's think of this problem like the Gambler's Ruin.

A bettor walks into the casino with 0 dollars. She makes \$1 wagers on coin flips; she wins \$1 on heads, and she loses \$1 on tails. She intends to make bets until she reaches a bankroll of \$$m$ (at which point she'll have seen $m$ more heads than tails). This is a very generous casino, so she is allowed to assume a very small amount of debt; she can borrow a maximum of \$$n$ from the casino (at which point she'll have seen $n$ more tails than heads). She will play until one of those conditions are met.

Let $M_t$ denote her bankroll after $t$ flips, and let $T$ be the number of flips required before she leaves. Note that $M_T$ is necessarily either $m$ or $-n$, but which one of those it is depends on the outcomes of the coin flips. Call $\mathbb P(M_T = m)$ by the name $p$. Since $M_t$ is the result of fair bets, it is a martingale. One can verify that the conditions of the Optional Stopping Theorem apply to $M_t$ and $T$, whence $\mathbb E[M_T] = \mathbb E[M_0] = 0$; that is, her expected bankroll when she leaves is the same as when she entered, because every bet was fair (and some technical conditions are satisfied). However, $$\mathbb E[M_T] = m \cdot p - n \cdot \mathbb (1-p)$$ so setting this equal to $0$ and solving for $p$ gives $p = \frac{n}{m+n}$. The intuition behind $p$ should somewhat clear; the starting point ($0$) is $n$ steps along the path of length $m+n$ from $-n$ to $m$.

It's worth pausing to point out that it's perhaps not at all clear why this is useful in answering the question you asked. Here's the magic: we'll observe a second martingale, $M'_t = M_t^2 - t$. You can easily see why this is a martingale; if $M_t = x$, then $M_t' = x^2 - t$, and $M_{t+1}'$ will be either $(x+1)^2-(t+1)$ or $(x-1)^2 - (t+1)$ with equal probability; you can verify that their average is $M_t'$.

Since $M_n'$ is a martingale, we can use the Optional Stopping Theorem up above again; note that $M_0' = 0$, whence $\mathbb E[M_T'] = 0$ as well. However, $$\mathbb E[M_T'] = \mathbb E[M_T^2] - \mathbb E[T] = m^2 p + n^2(1-p) - \mathbb E[T] = 0$$ and since we saw above that $p = \frac{n}{m + n}$, solving for $\mathbb E[T]$ gives the advertised $mn$.

I'm not sure if this will scratch your itch for a heuristic argument for the reasonableness of $mn$, but I like this method a lot and thought the algebra in the punch line may be illuminating.

Fair Coin

Let $e(h,t)$ be the expected duration to get $h$ more heads than tails or $t$ more tails than heads. We get the following relation: $$ e(h,t)=\overbrace{\tfrac12e(h-1,t+1)}^{\substack{\text{probability of a head}\\\text{times}\\\text{duration after a head}}}+\overbrace{\tfrac12e(h+1,t-1)}^{\substack{\text{probability of a tail}\\\text{times}\\\text{duration after a tail}}}+\overbrace{\ \quad1\vphantom{\tfrac12}\quad\ }^{\substack{\text{account}\vphantom{y}\\\text{for}\\\text{head/tail}}}\tag1 $$ If we let $f_{h+t}(h)=e(h,t)$. Then $(1)$ becomes a second order difference equation $$ f_n(h-1)-2f_n(h)+f_n(h+1)=-2\tag2 $$ Equation $(2)$ says that $f_n(h)$ is a degree $2$ polynomial in $h$ with with an $h^2$ coefficient of $-1$. Since $f_n(0)=f_n(n)=0$ we get $$ f_n(h)=h(n-h)\tag3 $$ Equation $(3)$ translates to $$ \bbox[5px,border:2px solid #C0A000]{e(h,t)=ht}\tag4 $$

Weighted Coin

Suppose the coin comes up heads with probability $p$ and tails with probability $1-p$. Equation $(1)$ becomes $$ e(h,t)=pe(h-1,t+1)+(1-p)e(h+1,t-1)+1\tag5 $$ which becomes the second order difference equation $$ pf_n(h-1)-f_n(h)+(1-p)f_n(h+1)=-1\tag6 $$ which, when written using the shift operator in $h$, $S_h$, is $$ (S_h-1)\left(S_h-\frac{p}{1-p}\right)f_n(h)=-\frac1{1-p}\tag7 $$ whose solution, given that $f_n(0)=f_n(n)=0$, is $$ f_n(h)=\frac{n}{1-2p}\,\frac{1-\left(\frac{p}{1-p}\right)^h}{1-\left(\frac{p}{1-p}\right)^n}+\frac{h}{2p-1}\tag8 $$ which translates to $$ \bbox[5px,border:2px solid #C0A000]{e(h,t)=\frac{t(1-p)^t\left[(1-p)^h-p^h\right]+hp^h\left[p^t-(1-p)^t\right]}{(1-2p)\left[(1-p)^{h+t}-p^{h+t}\right]}}\tag9 $$

Weighted Case Limits to the Fair Case

Simply plugging $p=\frac12$ into $(9)$ gives $\frac00$, not $(4)$.

To evaluate the limit of $(9)$ as $p\to\frac12$, set $p=\frac{1+\delta}2$, so $1-p=\frac{1-\delta}2$. Then, $(9)$ becomes $$ \begin{align} \hspace{-12pt}e(h,t) &=\frac{t(1-\delta)^t\left[(1-\delta)^h-(1+\delta)^h\right]+h(1+\delta)^h\left[(1+\delta)^t-(1-\delta)^t\right]}{\delta\left[(1+\delta)^{h+t}-(1-\delta)^{h+t}\right]}\\ &=\small\frac{t\left(1-t\delta+O\!\left(\delta^2\right)\right)\left(-2h\delta+O\!\left(\delta^3\right)\right)+h\left(1+h\delta+O\!\left(\delta^2\right)\right)\left(2t\delta+O\!\left(\delta^3\right)\right)}{2(h+t)\delta^2+O\!\left(\delta^4\right)}\\ &=\frac{2ht^2\delta^2+2h^2t\delta^2+O\!\left(\delta^3\right)}{2(h+t)\delta^2+O\!\left(\delta^4\right)}\\[3pt] &=\frac{2ht(h+t)+O(\delta)}{2(h+t)+O\!\left(\delta^2\right)}\\[6pt] &=ht+O(\delta)\tag{10} \end{align} $$ Thus, as $p\to\frac12$, $(10)$ shows that $(9)\to(4)$.