This boils down to the fact that the supercharges are represented in superspace by

$$ Q_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{m}\bar\theta)_{\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta^{\alpha}}\right)_{y} $$

$$ Q^{\dot\alpha}=\frac{\partial}{\partial\theta_{\dot\alpha}}-i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta_{\dot\alpha}}\right)_{y}-2i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial y^{m}} $$

And the D-terms and F-terms are written as fermionic integrals:

$$ \mathcal{L}_{D}=\int d^{2}\theta V(y,\theta),\,\,\,\,\,\,\,\,\,\,\, \mathcal{L}_{F}=\int d^2\theta d^{2}\bar\theta V(x,\theta,\bar\theta) $$

So if you act the supercharges in a D-term or a F-term, the first derivative of the supercharge, the $\theta$-derivative, will drop out because the fermionic integrals are only non-zero if it is saturated. This means that there will be only the $x$-derivatives, making a total derivative. This means that by doing a supersymmetric transformation in the Lagrangian we obtain a total derivative!, so the action is invariant under supersymmetry. Then we say that this Lagrangian is supersymmetric in a manifest way, since there is no need to check the invariance explicitly.

Now, for a given superfield, it is possible to write an action that is supersymmetric out of components that are not $F$-terms or $D$-terms, but this will be make the supersymmetry not manifest.

There is example for you. In pure spinor formalism for the second quantized superparticle in d=10 we can write the action as

$$ \int d^{10}x \langle \psi \,Q\psi\rangle $$

where $\psi(\lambda,\theta)$ is the pure spinor superfield, $Q=\lambda^{\alpha}D_{\alpha}$, and $\langle...\rangle$ is defined by picking just the $\langle\theta^5\lambda^3\rangle=1$, and zero otherwise. This means that it does not pick the last component of the superfields $\theta^{16}$. Nonetheless, this action describes linearized $d=10$ Super-Yang-Mills, so it is supersymmetric, although not in a manifest way.

Back to $d=4$, it is always possible to find an manifest supersymmetric formulation for a given non-manifest supersymmetric action. For large dimensions things start to be more complicated.

This boils down to the fact that the supercharges are represented in superspace by

$$ Q_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{m}\bar\theta)_{\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta^{\alpha}}\right)_{y} $$

$$ Q^{\dot\alpha}=\frac{\partial}{\partial\theta_{\dot\alpha}}-i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta_{\dot\alpha}}\right)_{y}-2i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial y^{m}} $$

And the D-terms and F-terms are written as fermionic integrals:

$$ \mathcal{L}_{D}=\int d^{2}\theta V(y,\theta),\,\,\,\,\,\,\,\,\,\,\, \mathcal{L}_{F}=\int d^2\theta d^{2}\bar\theta V(x,\theta,\bar\theta) $$

So if you act the supercharges in a D-term or a F-term, the first derivative of the supercharge, the $\theta$-derivative, will drop out because the fermionic integrals are only non-zero if it is saturated. This means that there will be only the $x$-derivatives, making a total derivative. This means that by doing a supersymmetric transformation in the Lagrangian we obtain a total derivative!, so the action is invariant under supersymmetry. Then we say that this Lagrangian is supersymmetric in a manifest way, since there is no need to check the invariance explicitly.

Now, for a given superfield, it is possible to write an action that is supersymmetric out of components that are not $F$-terms or $D$-terms, but this will be make the supersymmetry not manifest.

There is example for you. In pure spinor formalism for the second quantized superparticle in d=10 we can write the action as

$$ \int d^{10}x \langle \psi \,Q\psi\rangle $$

where $\psi(\lambda,\theta)$ is the pure spinor superfield, $Q=\lambda^{\alpha}D_{\alpha}$, and $\langle...\rangle$ is defined by picking just the $\langle\theta^5\lambda^3\rangle=1$, and zero otherwise. This means that it does not pick the last component of the superfields $\theta^{16}$. Nonetheless, this action describes linearized $d=10$ Super-Yang-Mills, so it is supersymmetric, although not in a manifest way.

Back to $d=4$, it is always possible to find an manifest supersymmetric formulation for a given non-manifest supersymmetric $N=1$ action. For large dimensions or extended supersymmetry things start to be more complicated, and the superspace should be generalized some how, e.g. harmonic superspace, pure spinors etc.

This boils down to the fact that the Supercharge is represented in superspace by something like

$$ Q_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}+i(\sigma^{m}\bar\theta)_{\alpha}\frac{\partial}{\partial x^{m}} $$

And in superspace, the D-terms and F-terms are written as fermionic integrals:

$$ \mathcal{L}_{D}=\int d^{2}\theta V(y,\theta),\,\,\,\,\,\,\,\,\,\,\, \mathcal{L}_{F}=\int d^2\theta d^{2}\bar\theta V(x,\theta,\bar\theta) $$

So if you act the supercharge in a D-term or a F-term, the first derivative, the $\theta$-derivative, will drop out because fermionic integrals are only non-zero if it is saturated. This means that there will be only the $x$-derivative, a total derivative. By doing a supersymmetric transformation in the Lagrangian we obtain a total derivative!, it means that the action is invariant under supersymmetry. Then we say that this Lagrangian is supersymmetric in a manifest way, since there is no need to check if it is invariant.

Now, for a given superfield, it is possible to write an action that is supersymmetric out of components that are not $F$-terms or $D$-terms, but this will be make the supersymmetry not manifest. In pure spinor formalism for the second quantized superparticle in d=10 we can write the action as

$$ \int d^{10}x \langle \psi \,Q\psi\rangle $$

where $\psi(\lambda,\theta)$ is the pure spinor superfield, $Q=\lambda^{\alpha}D_{\alpha}$, and $\langle...\rangle$ is defined by picking just the $\langle\theta^5\lambda^3\rangle=1$, and zero otherwise. This means that it does not pick the last component of the superfields $\theta^{16}$. Nonetheless, this action describes linearized $d=10$ Super-Yang-Mills, so it is supersymmetric, although not in a manifest way.

This boils down to the fact that the supercharges are represented in superspace by

$$ Q_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{m}\bar\theta)_{\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta^{\alpha}}\right)_{y} $$

$$ Q^{\dot\alpha}=\frac{\partial}{\partial\theta_{\dot\alpha}}-i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial x^{m}}=\left(\frac{\partial}{\partial\theta_{\dot\alpha}}\right)_{y}-2i(\bar\sigma^{m}\theta)^{\dot\alpha}\frac{\partial}{\partial y^{m}} $$

And the D-terms and F-terms are written as fermionic integrals:

$$ \mathcal{L}_{D}=\int d^{2}\theta V(y,\theta),\,\,\,\,\,\,\,\,\,\,\, \mathcal{L}_{F}=\int d^2\theta d^{2}\bar\theta V(x,\theta,\bar\theta) $$

So if you act the supercharges in a D-term or a F-term, the first derivative of the supercharge, the $\theta$-derivative, will drop out because the fermionic integrals are only non-zero if it is saturated. This means that there will be only the $x$-derivatives, making a total derivative. This means that by doing a supersymmetric transformation in the Lagrangian we obtain a total derivative!, so the action is invariant under supersymmetry. Then we say that this Lagrangian is supersymmetric in a manifest way, since there is no need to check the invariance explicitly.

Now, for a given superfield, it is possible to write an action that is supersymmetric out of components that are not $F$-terms or $D$-terms, but this will be make the supersymmetry not manifest.

There is example for you. In pure spinor formalism for the second quantized superparticle in d=10 we can write the action as

$$ \int d^{10}x \langle \psi \,Q\psi\rangle $$

where $\psi(\lambda,\theta)$ is the pure spinor superfield, $Q=\lambda^{\alpha}D_{\alpha}$, and $\langle...\rangle$ is defined by picking just the $\langle\theta^5\lambda^3\rangle=1$, and zero otherwise. This means that it does not pick the last component of the superfields $\theta^{16}$. Nonetheless, this action describes linearized $d=10$ Super-Yang-Mills, so it is supersymmetric, although not in a manifest way.

Back to $d=4$, it is always possible to find an manifest supersymmetric formulation for a given non-manifest supersymmetric action. For large dimensions things start to be more complicated.