One easy way to understand interpolation in the time domain by zero-padding in the frequency domain is to realize that all interpolated sequences can be derived from sampling a single periodic *continuous-time* function, defined by the DFT coefficients $X[k]$, which are interpreted as (scaled) Fourier coefficients of that periodic continuous-time function $x_c(t)$. For odd $N$ we have

$$x_c(t)=\frac{1}{N}\sum_{k=-(N-1)/2}^{(N-1)/2}X[k]e^{j2\pi 
kt/N}\tag{1}$$

and for even $N$ (your example) you get

$$x_c(t)=\frac{1}{N}\sum_{k=-N/2}^{N/2}\tilde{X}[k]e^{j2\pi kt/N}\tag{2}$$

where $\tilde{X}[k]$ is obtained from $X[k]$ by splitting the bin at Nyquist (index $N/2$):

$$\tilde{X}[k]=\big[X[0],\ldots,X[N/2-1],0.5X[N/2],\\0.5X[N/2],X[N/2+1],\ldots,X[N-1]\big]$$

where we assume periodicity with period $N+1$ (due to splitting of the Nyquist bin): $\tilde{X}[k]=X[k+N+1]$, so $\tilde{X}[-N/2]=\tilde{X}[N/2+1]=0.5X[N/2]$.

Note that for real-valued $x[n]$, $x_c(t)$ defined by $(1)$ or $(2)$ is real-valued. Also note that regardless of the interpolation factor, all interpolated discrete-time sequences are samples of $x_c(t)$. So the blue curves in your question do not make much sense, or they at least don't help with understanding what's going on.

For a given length $M$ of the desired interpolated sequence ($M>N$), the interpolated sequence obtained by IDFT from zero-padding in the frequency domain can be written in terms of a sampled version of $x_c(t)$:

$$\hat{x}[m]=x_c\left(\frac{mN}{M}\right)=\frac{M}{N}\textrm{IDFT}_M\{X_{ZP}[k]\}\tag{3}$$

where $X_{ZP}[k]$ is a zero-padded version of $X[k]$ ($N$ odd) or $\tilde{X}[k]$ ($N$ even), respectively. The amplitude scaling in your plots is due to the factor $M/N$ in $(3)$ that you probably forgot to include in your computations.